Fundamental definite integration Questions in English

(D) Let $f(x) = \prod_{r=1}^{n} (x+r)$.
Then $\ln(f(x)) = \sum_{k=1}^{n} \ln(x+k)$.
Differentiating both sides with respect to $x$,we get $\frac{f'(x)}{f(x)} = \sum_{k=1}^{n} \frac{1}{x+k}$.
Thus,$f'(x) = f(x) \sum_{k=1}^{n} \frac{1}{x+k}$.
The integral becomes $\int_{0}^{1} f'(x) dx = [f(x)]_{0}^{1}$.
$f(1) = \prod_{r=1}^{n} (1+r) = 2 \cdot 3 \cdot \dots \cdot (n+1) = (n+1)!$.
$f(0) = \prod_{r=1}^{n} (0+r) = 1 \cdot 2 \cdot \dots \cdot n = n!$.
Therefore,the value of the integral is $(n+1)! - n! = n!(n+1-1) = n \cdot n!$.

(D) First,evaluate the integrals:
$\int_{0}^{x} dz = [z]_{0}^{x} = x$
$\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
Substitute these into the inequality:
$\sqrt{5x - 6 - x^2} + \frac{\pi}{2} x > x \left( \frac{\pi}{2} \right)$
$\sqrt{5x - 6 - x^2} + \frac{\pi x}{2} > \frac{\pi x}{2}$
Subtract $\frac{\pi x}{2}$ from both sides:
$\sqrt{5x - 6 - x^2} > 0$
For the square root to be defined and greater than zero,the expression inside must be strictly positive:
$5x - 6 - x^2 > 0$
$-(x^2 - 5x + 6) > 0$
$x^2 - 5x + 6 < 0$
$(x - 2)(x - 3) < 0$
Thus,the solution set is $x \in (2, 3)$.

(B) Let $I = \int_1^e \left( \frac{\tan^{-1}x}{x} + \frac{\ln x}{1+x^2} \right) dx$.
We can split this into two integrals: $I = \int_1^e \frac{\tan^{-1}x}{x} dx + \int_1^e \frac{\ln x}{1+x^2} dx$.
Applying integration by parts to the first integral $\int_1^e \tan^{-1}x \cdot \frac{1}{x} dx$:
Let $u = \tan^{-1}x$ and $dv = \frac{1}{x} dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \ln x$.
Using the formula $\int u dv = uv - \int v du$:
$\int_1^e \frac{\tan^{-1}x}{x} dx = [\tan^{-1}x \cdot \ln x]_1^e - \int_1^e \frac{\ln x}{1+x^2} dx$.
Substituting this back into the expression for $I$:
$I = \left( [\tan^{-1}x \cdot \ln x]_1^e - \int_1^e \frac{\ln x}{1+x^2} dx \right) + \int_1^e \frac{\ln x}{1+x^2} dx$.
The integral terms cancel out:
$I = [\tan^{-1}x \cdot \ln x]_1^e$.
Evaluating the limits:
$I = (\tan^{-1}e \cdot \ln e) - (\tan^{-1}1 \cdot \ln 1)$.
Since $\ln e = 1$ and $\ln 1 = 0$:
$I = \tan^{-1}e \cdot 1 - 0 = \tan^{-1}e$.

(D) The given integral is $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \log_e(\sin x + \cos x) \, dx$.
According to the property of definite integrals,$\int_{a}^{a} f(x) \, dx = 0$ for any finite value $a$.
Since the lower limit and the upper limit of the integral are the same (i.e.,$\frac{\pi}{4}$),the value of the integral is $0$.

(A) We are given the integral $I = \int_{0}^{9} [\sqrt{x} + 2] \, dx$.
Using the property of the Greatest Integer Function $[x + n] = [x] + n$ for any integer $n$,we have $[\sqrt{x} + 2] = [\sqrt{x}] + 2$.
Thus,$I = \int_{0}^{9} [\sqrt{x}] \, dx + \int_{0}^{9} 2 \, dx$.
Evaluating the second part: $\int_{0}^{9} 2 \, dx = 2[x]_{0}^{9} = 2(9 - 0) = 18$.
For the first part,we split the interval $[0, 9]$ based on the values where $\sqrt{x}$ is an integer:
$\int_{0}^{9} [\sqrt{x}] \, dx = \int_{0}^{1} [\sqrt{x}] \, dx + \int_{1}^{4} [\sqrt{x}] \, dx + \int_{4}^{9} [\sqrt{x}] \, dx$.
In $[0, 1)$,$[\sqrt{x}] = 0$.
In $[1, 4)$,$[\sqrt{x}] = 1$.
In $[4, 9)$,$[\sqrt{x}] = 2$.
So,$\int_{0}^{9} [\sqrt{x}] \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{4} 1 \, dx + \int_{4}^{9} 2 \, dx$.
$= 0 + (4 - 1) + 2(9 - 4) = 3 + 2(5) = 3 + 10 = 13$.
Finally,$I = 13 + 18 = 31$.

(A) Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.
Squaring both sides,we get $y^2 = x + y$,which implies $y^2 - y - x = 0$.
Solving for $y$ using the quadratic formula,$y = \frac{1 + \sqrt{1 + 4x}}{2}$ (since $y > 0$).
Now,we need to evaluate $I = \int_{0}^{2} \frac{1 + \sqrt{1 + 4x}}{2} \, dx$.
$I = \frac{1}{2} \int_{0}^{2} 1 \, dx + \frac{1}{2} \int_{0}^{2} (1 + 4x)^{1/2} \, dx$.
$I = \frac{1}{2} [x]_{0}^{2} + \frac{1}{2} \left[ \frac{(1 + 4x)^{3/2}}{3/2 \cdot 4} \right]_{0}^{2}$.
$I = \frac{1}{2} (2) + \frac{1}{12} [(1 + 8)^{3/2} - (1 + 0)^{3/2}]$.
$I = 1 + \frac{1}{12} [27 - 1] = 1 + \frac{26}{12} = 1 + \frac{13}{6} = \frac{19}{6} \approx 3.166$.
The Greatest Integer Function $[I] = [3.166] = 3$.

(C) Given the equation $f(x) = \int\limits_1^x {tf(t)dt}$.
By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$,we get $f'(x) = xf(x)$.
Also,substituting $x = 1$,we get $f(1) = \int\limits_1^1 {tf(t)dt} = 0$.
Now,$\frac{f'(x)}{f(x)} = x$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = \frac{x^2}{2} + C$.
This implies $f(x) = A e^{x^2/2}$.
Using the condition $f(1) = 0$,we have $A e^{1/2} = 0$,which implies $A = 0$.
Therefore,$f(x) = 0$ for all $x \in R$.
Since $f(x) = 0$,the integral $\int\limits_{-3}^3 f(x) dx = \int\limits_{-3}^3 0 dx = 0$.
Thus,the correct statement is $\int\limits_{-3}^3 f(x) dx = 0$.

(B) Given $y = \sqrt{\sec x + y}$.
Squaring both sides,we get $y^2 = \sec x + y$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = \sec x \tan x + \frac{dy}{dx}$.
Rearranging the terms,we have $(2y - 1) \frac{dy}{dx} = \sec x \tan x$.
Now,we need to evaluate the integral $I = \int_{0}^{\pi/3} (2y - 1) \frac{dy}{dx} \, dx$.
Substituting the expression derived above,$I = \int_{0}^{\pi/3} \sec x \tan x \, dx$.
The integral of $\sec x \tan x$ is $\sec x$.
Thus,$I = [\sec x]_{0}^{\pi/3} = \sec(\pi/3) - \sec(0) = 2 - 1 = 1$.

(B) We need to evaluate $I = \int_{-1}^{2} \left[ \frac{[x]}{1 + x^2} \right] dx$.
We split the integral based on the intervals of $[x]$:
For $-1 \le x < 0$,$[x] = -1$. Thus,the integrand is $\left[ \frac{-1}{1 + x^2} \right]$. Since $0 < 1 + x^2 \le 2$,we have $-1 \le \frac{-1}{1 + x^2} < -0.5$. The greatest integer of this value is $-1$.
For $0 \le x < 1$,$[x] = 0$. Thus,the integrand is $\left[ \frac{0}{1 + x^2} \right] = [0] = 0$.
For $1 \le x < 2$,$[x] = 1$. Thus,the integrand is $\left[ \frac{1}{1 + x^2} \right]$. Since $2 < 1 + x^2 \le 5$,we have $0.2 \le \frac{1}{1 + x^2} < 0.5$. The greatest integer of this value is $0$.
Therefore,$I = \int_{-1}^{0} (-1) dx + \int_{0}^{1} 0 dx + \int_{1}^{2} 0 dx = [-x]_{-1}^{0} + 0 + 0 = (0 - 1) = -1$.

(A) Let $f(x) = ax^2 + bx + c$.
Then,$\int_{0}^{1} f(x) dx = \int_{0}^{1} (ax^2 + bx + c) dx = \left[ \frac{ax^3}{3} + \frac{bx^2}{2} + cx \right]_0^1 = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Now,calculate the values of $f(x)$ at the given points:
$f(0) = c$
$f\left(\frac{1}{2}\right) = a\left(\frac{1}{4}\right) + b\left(\frac{1}{2}\right) + c = \frac{a + 2b + 4c}{4}$
$f(1) = a + b + c$
Consider the expression $\frac{1}{6}\left( f(0) + 4f\left(\frac{1}{2}\right) + f(1) \right)$:
$= \frac{1}{6}\left( c + 4\left(\frac{a + 2b + 4c}{4}\right) + (a + b + c) \right)$
$= \frac{1}{6}\left( c + a + 2b + 4c + a + b + c \right)$
$= \frac{1}{6}\left( 2a + 3b + 6c \right)$.
This matches the integral value. Thus,the correct option is $A$.

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} dx$.
We can rewrite the integrand as:
$\frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} = 2x \sqrt{\sin x} + \frac{x^2 \cos x}{2\sqrt{\sin x}}$.
Notice that this is the derivative of the product $f(x) = x^2 \sqrt{\sin x}$.
Using the product rule: $\frac{d}{dx}(x^2 \sqrt{\sin x}) = 2x \sqrt{\sin x} + x^2 \cdot \frac{1}{2\sqrt{\sin x}} \cdot \cos x = 2x \sqrt{\sin x} + \frac{x^2 \cos x}{2\sqrt{\sin x}}$.
Thus,the integral becomes:
$I = \int_{0}^{\frac{\pi}{2}} \frac{d}{dx}(x^2 \sqrt{\sin x}) dx$.
By the Fundamental Theorem of Calculus:
$I = [x^2 \sqrt{\sin x}]_{0}^{\frac{\pi}{2}} = ((\frac{\pi}{2})^2 \sqrt{\sin(\frac{\pi}{2})}) - (0^2 \sqrt{\sin(0)}) = \frac{\pi^2}{4} \cdot 1 - 0 = \frac{\pi^2}{4}$.

(D) Let $I = \sum\limits_{r = 2}^{16} {\int_r^{r + 1} {\frac{{dx}}{{(2r - x)(2r + 2 - x)}}} }$.
For each integral,let $u = x - (2r + 1)$. Then $du = dx$.
When $x = r$,$u = r - 2r - 1 = -r - 1$.
When $x = r + 1$,$u = r + 1 - 2r - 1 = -r$.
The denominator becomes $(2r - (u + 2r + 1))(2r + 2 - (u + 2r + 1)) = (-u - 1)(1 - u) = (u + 1)(u - 1) = u^2 - 1$.
Thus,$\int_r^{r+1} \frac{dx}{(2r-x)(2r+2-x)} = \int_{-r-1}^{-r} \frac{du}{u^2 - 1} = \left[ \frac{1}{2} \ln \left| \frac{u-1}{u+1} \right| \right]_{-r-1}^{-r}$.
$= \frac{1}{2} \left( \ln \left| \frac{-r-1}{-r+1} \right| - \ln \left| \frac{-r-1-1}{-r-1+1} \right| \right) = \frac{1}{2} \left( \ln \left( \frac{r+1}{r-1} \right) - \ln \left( \frac{r+2}{r} \right) \right)$.
Summing from $r=2$ to $16$: $\frac{1}{2} \sum_{r=2}^{16} \left( \ln \frac{r+1}{r-1} - \ln \frac{r+2}{r} \right)$.
This is a telescoping sum: $\frac{1}{2} [(\ln \frac{3}{1} - \ln \frac{4}{2}) + (\ln \frac{4}{2} - \ln \frac{5}{3}) + \dots + (\ln \frac{17}{15} - \ln \frac{18}{16})]$.
$= \frac{1}{2} (\ln 3 - \ln \frac{18}{16}) = \frac{1}{2} \ln \left( \frac{3 \times 16}{18} \right) = \frac{1}{2} \ln \left( \frac{48}{18} \right) = \frac{1}{2} \ln \left( \frac{8}{3} \right) = \ln \sqrt{\frac{8}{3}} = \ln \left( \frac{2\sqrt{2}}{\sqrt{3}} \right)$.

(D) We know that $\sin 3y = 3\sin y - 4\sin^3 y$.
Substituting this into the integrand:
$\frac{1 + 3\sin y - 4\sin^3 y}{1 + 2\sin y}$.
Using polynomial division or factoring:
$1 + 3\sin y - 4\sin^3 y = (1 + 2\sin y)(1 + \sin y - 2\sin^2 y)$.
Thus,the integrand simplifies to $1 + \sin y - 2\sin^2 y$.
Now,integrate with respect to $y$ from $0$ to $\frac{\pi}{2}$:
$\int_{0}^{\frac{\pi}{2}} (1 + \sin y - 2\sin^2 y) dy = \int_{0}^{\frac{\pi}{2}} (1 + \sin y - (1 - \cos 2y)) dy$.
$= \int_{0}^{\frac{\pi}{2}} (\sin y + \cos 2y) dy$.
$= [-\cos y + \frac{1}{2}\sin 2y]_{0}^{\frac{\pi}{2}}$.
$= (-\cos \frac{\pi}{2} + \frac{1}{2}\sin \pi) - (-\cos 0 + \frac{1}{2}\sin 0)$.
$= (0 + 0) - (-1 + 0) = 1$.

(B) We evaluate the integral using integration by parts: $\int u dv = uv - \int v du$. Let $u = t$ and $dv = \sin(x + \pi t) dt$. Then $du = dt$ and $v = -\frac{1}{\pi} \cos(x + \pi t)$.
$f(x) = \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 + \frac{1}{\pi} \int_0^1 \cos(x + \pi t) dt$
$f(x) = \left( -\frac{1}{\pi} \cos(x + \pi) - 0 \right) + \frac{1}{\pi} \left[ \frac{1}{\pi} \sin(x + \pi t) \right]_0^1$
Since $\cos(x + \pi) = -\cos x$,we have $f(x) = \frac{1}{\pi} \cos x + \frac{1}{\pi^2} (\sin(x + \pi) - \sin x)$.
Since $\sin(x + \pi) = -\sin x$,we have $f(x) = \frac{1}{\pi} \cos x - \frac{2}{\pi^2} \sin x$.
The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$.
Here $a = \frac{1}{\pi}$ and $b = -\frac{2}{\pi^2}$.
$f(x)_{\max} = \sqrt{\left(\frac{1}{\pi}\right)^2 + \left(-\frac{2}{\pi^2}\right)^2} = \sqrt{\frac{1}{\pi^2} + \frac{4}{\pi^4}} = \sqrt{\frac{\pi^2 + 4}{\pi^4}} = \frac{\sqrt{\pi^2 + 4}}{\pi^2}$.

(A) Let $I = \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$.
Expanding $(1-x)^{4} = 1 - 4x + 6x^{2} - 4x^{3} + x^{4}$.
So,$I = \int_{0}^{1} \frac{x^{4}(1 - 4x + 6x^{2} - 4x^{3} + x^{4})}{1+x^{2}} dx$.
Performing polynomial division of $x^{4}(x^{4} - 4x^{3} + 6x^{2} - 4x + 1)$ by $(x^{2} + 1)$:
$x^{8} - 4x^{7} + 6x^{6} - 4x^{5} + x^{4} = (x^{2} + 1)(x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4) - 4$.
Thus,$\frac{x^{4}(1-x)^{4}}{1+x^{2}} = x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4 - \frac{4}{1+x^{2}}$.
Integrating term by term from $0$ to $1$:
$I = \int_{0}^{1} (x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4) dx - 4 \int_{0}^{1} \frac{1}{1+x^{2}} dx$.
$I = [\frac{x^{7}}{7} - \frac{4x^{6}}{6} + \frac{5x^{5}}{5} - \frac{4x^{3}}{3} + 4x]_{0}^{1} - 4[\tan^{-1}(x)]_{0}^{1}$.
$I = (\frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4) - 4(\frac{\pi}{4})$.
$I = (\frac{3 - 14 + 21 - 28 + 84}{21}) - \pi = \frac{66}{21} - \pi = \frac{22}{7} - \pi$.

(A) We need to evaluate the integral $\int_{0}^{2} f(x) dx$. Since the function is defined piecewise,we split the integral at $x=1$:
$\int_{0}^{2} f(x) dx = \int_{0}^{1} \sqrt{1-x} dx + \int_{1}^{2} (7x-6)^{-1/3} dx$.
For the first part,let $u = 1-x$,then $du = -dx$. When $x=0, u=1$; when $x=1, u=0$:
$\int_{0}^{1} (1-x)^{1/2} dx = -\int_{1}^{0} u^{1/2} du = \int_{0}^{1} u^{1/2} du = [\frac{2}{3} u^{3/2}]_{0}^{1} = \frac{2}{3}$.
For the second part,let $v = 7x-6$,then $dv = 7 dx$,so $dx = \frac{1}{7} dv$. When $x=1, v=1$; when $x=2, v=8$:
$\int_{1}^{2} (7x-6)^{-1/3} dx = \frac{1}{7} \int_{1}^{8} v^{-1/3} dv = \frac{1}{7} [\frac{3}{2} v^{2/3}]_{1}^{8} = \frac{1}{7} \cdot \frac{3}{2} (8^{2/3} - 1^{2/3}) = \frac{3}{14} (4 - 1) = \frac{3}{14} \cdot 3 = \frac{9}{14}$.
Adding the two parts: $\frac{2}{3} + \frac{9}{14} = \frac{28 + 27}{42} = \frac{55}{42}$.

(D) Let $I = \int_{0}^{1} e^{x e^{x}} (1 + x e^{x}) dx$.
Consider the substitution $u = x e^{x}$.
Then,$du = (1 \cdot e^{x} + x \cdot e^{x}) dx = e^{x}(1 + x) dx$. This does not match the integrand directly.
Let us re-examine the integral: $\int_{0}^{1} e^{x e^{x}} (1 + x e^{x}) dx$.
Let $f(x) = x e^{x}$. Then $f'(x) = e^{x} + x e^{x} = e^{x}(1 + x)$. This is not the term $(1 + x e^{x})$.
Wait,let us check the derivative of $e^{x e^{x}}$.
$\frac{d}{dx} (e^{x e^{x}}) = e^{x e^{x}} \cdot \frac{d}{dx} (x e^{x}) = e^{x e^{x}} (e^{x} + x e^{x}) = e^{x e^{x}} e^{x} (1 + x)$.
This does not match. Let us re-evaluate the integral $\int_{0}^{1} e^{e^{x}} (1 + x e^{x}) dx$ as provided in the original prompt.
Let $u = e^{x}$. Then $du = e^{x} dx$,so $dx = \frac{du}{u}$.
When $x=0, u=1$. When $x=1, u=e$.
$I = \int_{1}^{e} e^{u} (1 + (\ln u) u) \frac{du}{u} = \int_{1}^{e} e^{u} (\frac{1}{u} + \ln u) du$.
This integral does not have a simple elementary antiderivative.
However,if the question was $\int_{0}^{1} e^{x e^{x}} (1 + x e^{x}) dx$ is not standard,let us check $\int_{0}^{1} \frac{d}{dx} (e^{x e^{x}}) dx = [e^{x e^{x}}]_{0}^{1} = e^{e} - e^{0} = e^{e} - 1$.
Thus,the correct value is $e^{e} - 1$.

(A) Given $\phi(x) = e^{x} \int_{0}^{1} e^{t} \phi(t) dt + x$. Let $C = \int_{0}^{1} e^{t} \phi(t) dt$. Then $\phi(x) = C e^{x} + x$.
Substituting $\phi(t) = C e^{t} + t$ into the integral:
$C = \int_{0}^{1} e^{t} (C e^{t} + t) dt = \int_{0}^{1} (C e^{2t} + t e^{t}) dt$.
Evaluating the integral:
$C = C \left[ \frac{e^{2t}}{2} \right]_{0}^{1} + \left[ t e^{t} - e^{t} \right]_{0}^{1} = C \left( \frac{e^{2} - 1}{2} \right) + (e - e) - (0 - 1) = C \left( \frac{e^{2} - 1}{2} \right) + 1$.
$C \left( 1 - \frac{e^{2} - 1}{2} \right) = 1 \Rightarrow C \left( \frac{2 - e^{2} + 1}{2} \right) = 1 \Rightarrow C \left( \frac{3 - e^{2}}{2} \right) = 1 \Rightarrow C = \frac{2}{3 - e^{2}}$.
Thus,$\phi(x) = \frac{2}{3 - e^{2}} e^{x} + x$.
Now,$\phi(\ln(e^{2} - 3)) = \frac{2}{3 - e^{2}} e^{\ln(e^{2} - 3)} + \ln(e^{2} - 3) = \frac{2}{3 - e^{2}} (e^{2} - 3) + \ln(e^{2} - 3) = -2 + \ln(e^{2} - 3)$.

(B) For $0 < x < \frac{\pi}{6}$,we know that $\cos x > x$ (since $\cos x$ is decreasing and $x$ is increasing,and at $x=0$,$\cos 0 = 1 > 0$).
Thus,$\frac{\cos x}{x} > 1$.
Integrating both sides from $0$ to $\frac{\pi}{6}$:
$I = \int_{0}^{\pi/6} \frac{\cos x}{x} dx > \int_{0}^{\pi/6} 1 dx = \frac{\pi}{6}$.
So,$I > \frac{\pi}{6}$.
For $\frac{\pi}{3} < x < \frac{\pi}{2}$,we know that $\cos x < x$ (since $\cos(\pi/3) = 0.5 < \pi/3 \approx 1.047$).
Thus,$\frac{\cos x}{x} < 1$.
Integrating both sides from $\frac{\pi}{3}$ to $\frac{\pi}{2}$:
$J = \int_{\pi/3}^{\pi/2} \frac{\cos x}{x} dx < \int_{\pi/3}^{\pi/2} 1 dx = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.
So,$J < \frac{\pi}{6}$.
Therefore,the correct option is $I > \frac{\pi}{6}$ and $J < \frac{\pi}{6}$.

(C) Let $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.
Substitute $x = \alpha \cos^2 t + \beta \sin^2 t$.
Then $dx = 2(\beta - \alpha) \sin t \cos t dt$.
When $x = \alpha$,$t = 0$. When $x = \beta$,$t = \frac{\pi}{2}$.
$x - \alpha = (\beta - \alpha) \sin^2 t$ and $\beta - x = (\beta - \alpha) \cos^2 t$.
So,$\sqrt{\frac{x-\alpha}{\beta-x}} = \sqrt{\frac{(\beta-\alpha)\sin^2 t}{(\beta-\alpha)\cos^2 t}} = \tan t$.
$I = \int_{0}^{\pi/2} \tan t \cdot 2(\beta - \alpha) \sin t \cos t dt = 2(\beta - \alpha) \int_{0}^{\pi/2} \sin^2 t dt$.
Using the identity $\sin^2 t = \frac{1 - \cos 2t}{2}$,we get:
$I = 2(\beta - \alpha) \int_{0}^{\pi/2} \frac{1 - \cos 2t}{2} dt = (\beta - \alpha) [t - \frac{\sin 2t}{2}]_{0}^{\pi/2}$.
$I = (\beta - \alpha) [\frac{\pi}{2} - 0] = \frac{\pi}{2}(\beta - \alpha)$.

(D) Consider the function $g(t) = e^{-t^2}$. For $t \in (0, x)$,the function $g(t)$ is strictly decreasing because $g'(t) = -2t e^{-t^2} < 0$ for $t > 0$.
Since $g(t)$ is a strictly decreasing concave function on $(0, x)$,the area under the curve $y = e^{-t^2}$ from $0$ to $x$ is greater than the area of the trapezoid formed by the points $(0, 0), (x, 0), (x, e^{-x^2}),$ and $(0, 1)$.
The area of this trapezoid is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height}) = \frac{1}{2} (1 + e^{-x^2}) \times x$.
Therefore,$f(x) = \int_0^x e^{-t^2} dt > \frac{x}{2} (1 + e^{-x^2})$.

(D) Let $I = \int_{1}^{e} (x+1) e^{x} \ln x \, dx$.
We know that $\int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C$.
Rewrite the integrand as:
$(x+1) e^{x} \ln x = e^{x} (x \ln x + \ln x) = e^{x} (x \ln x + \ln x + 1 - 1) = e^{x} ((x \ln x) + (\ln x + 1) - 1)$.
This does not immediately fit the form. Let us use integration by parts.
Let $u = \ln x$ and $dv = (x+1)e^{x} dx$.
Then $du = \frac{1}{x} dx$ and $v = \int (x+1)e^{x} dx = x e^{x}$.
Using $\int u \, dv = uv - \int v \, du$:
$I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} x e^{x} \cdot \frac{1}{x} \, dx$
$I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} e^{x} \, dx$
$I = [x e^{x} \ln x - e^{x}]_{1}^{e}$
$I = (e \cdot e^{e} \cdot \ln e - e^{e}) - (1 \cdot e^{1} \cdot \ln 1 - e^{1})$
$I = (e \cdot e^{e} \cdot 1 - e^{e}) - (0 - e)$
$I = e^{e+1} - e^{e} + e = e^{e}(e-1) + e$.

(C) We know that $\{ \sqrt{x} \} = \sqrt{x} - [\sqrt{x}]$.
So,the integral becomes $\int_{0}^{4} (\sqrt{x} - [\sqrt{x}]) dx = \int_{0}^{4} \sqrt{x} dx - \int_{0}^{4} [\sqrt{x}] dx$.
First,calculate $\int_{0}^{4} \sqrt{x} dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = \frac{2}{3} (4^{3/2} - 0) = \frac{2}{3} \times 8 = \frac{16}{3}$.
Next,evaluate $\int_{0}^{4} [\sqrt{x}] dx$. The value of $[\sqrt{x}]$ changes at $x = 1, 4, 9, \dots$:
For $0 \le x < 1$,$[\sqrt{x}] = 0$.
For $1 \le x < 4$,$[\sqrt{x}] = 1$.
Thus,$\int_{0}^{4} [\sqrt{x}] dx = \int_{0}^{1} 0 dx + \int_{1}^{4} 1 dx = 0 + [x]_{1}^{4} = 4 - 1 = 3$.
Finally,the value of the integral is $\frac{16}{3} - 3 = \frac{16 - 9}{3} = \frac{7}{3}$.

(C) We know that $[x + n] = [x] + n$ for any integer $n$.
Given the expression $[x + [x + [x]]]$,we can simplify it step by step:
$[x + [x + [x]]] = [x + [x] + [x]] = [x + 2[x]] = [x] + 2[x] = 3[x]$.
Now,we evaluate the integral:
$\int_{-1}^{1} 3[x] \, dx = 3 \int_{-1}^{1} [x] \, dx$.
Splitting the integral at $x = 0$:
$3 \left( \int_{-1}^{0} [x] \, dx + \int_{0}^{1} [x] \, dx \right)$.
For $x \in [-1, 0)$,$[x] = -1$.
For $x \in [0, 1)$,$[x] = 0$.
So,$3 \left( \int_{-1}^{0} (-1) \, dx + \int_{0}^{1} (0) \, dx \right) = 3 \left( [-x]_{-1}^{0} + 0 \right) = 3(-1) = -3$.

(C) Given that $\int_{n}^{n+1} f(x) dx = n^2 + n$.
We need to evaluate $\int_{-3}^{3} f(x) dx$.
Using the property of definite integrals,we can write:
$\int_{-3}^{3} f(x) dx = \int_{-3}^{-2} f(x) dx + \int_{-2}^{-1} f(x) dx + \int_{-1}^{0} f(x) dx + \int_{0}^{1} f(x) dx + \int_{1}^{2} f(x) dx + \int_{2}^{3} f(x) dx$.
Now,substitute $n = -3, -2, -1, 0, 1, 2$ into the given formula $n^2 + n$:
For $n = -3$: $(-3)^2 + (-3) = 9 - 3 = 6$.
For $n = -2$: $(-2)^2 + (-2) = 4 - 2 = 2$.
For $n = -1$: $(-1)^2 + (-1) = 1 - 1 = 0$.
For $n = 0$: $(0)^2 + 0 = 0$.
For $n = 1$: $(1)^2 + 1 = 2$.
For $n = 2$: $(2)^2 + 2 = 6$.
Summing these values: $6 + 2 + 0 + 0 + 2 + 6 = 16$.

(B) We define the function $f(x) = |x \sin(\pi x)|$. The expression $x \sin(\pi x)$ changes sign at $x = 0$ and $x = 1$.
For $x \in [-1, 0]$,$x \leq 0$ and $\sin(\pi x) \leq 0$,so $x \sin(\pi x) \geq 0$.
For $x \in [0, 1]$,$x \geq 0$ and $\sin(\pi x) \geq 0$,so $x \sin(\pi x) \geq 0$.
For $x \in [1, 3/2]$,$x > 0$ and $\sin(\pi x) < 0$,so $x \sin(\pi x) < 0$.
Thus,$|x \sin(\pi x)| = x \sin(\pi x)$ for $x \in [-1, 1]$ and $-x \sin(\pi x)$ for $x \in [1, 3/2]$.
$\int_{-1}^{3/2} |x \sin(\pi x)| dx = \int_{-1}^{1} x \sin(\pi x) dx - \int_{1}^{3/2} x \sin(\pi x) dx$.
Using integration by parts,$\int x \sin(\pi x) dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} + C$.
Evaluating the first part:
$\int_{-1}^{1} x \sin(\pi x) dx = \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{-1}^{1} = \left( -\frac{1 \cdot (-1)}{\pi} + 0 \right) - \left( -\frac{(-1) \cdot (-1)}{\pi} + 0 \right) = \frac{1}{\pi} - (-\frac{1}{\pi}) = \frac{2}{\pi}$.
Evaluating the second part:
$\int_{1}^{3/2} x \sin(\pi x) dx = \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = \left( 0 + \frac{\sin(3\pi/2)}{\pi^2} \right) - \left( -\frac{1 \cdot (-1)}{\pi} + 0 \right) = -\frac{1}{\pi^2} - \frac{1}{\pi}$.
Total integral = $\frac{2}{\pi} - (-\frac{1}{\pi^2} - \frac{1}{\pi}) = \frac{2}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

(C) Let $I = \int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$.
Substitute $u = g(x)$,then $du = g'(x) dx$.
When $x = 1$,$u = g(1)$.
When $x = 2$,$u = g(2)$.
Since $g(1) = g(2) = t$,the integral becomes:
$I = \int_{g(1)}^{g(2)} [f(u)]^{-1} f'(u) du$
$I = \int_{t}^{t} \frac{f'(u)}{f(u)} du$
By the property of definite integrals,if the upper and lower limits are equal,the integral is $0$.
$I = [\ln|f(u)|]_{t}^{t} = \ln|f(t)| - \ln|f(t)| = 0$.

(A) We know that $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.
Substituting this into the integral:
$I = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x}{(\frac{2}{\sin 2x})^3} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x \sin^3 2x}{8} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \cos 2x \sin^3 2x dx$.
Let $u = \sin 2x$,then $du = 2 \cos 2x dx$,so $\cos 2x dx = \frac{du}{2}$.
When $x = \frac{\pi}{12}$,$u = \sin(\frac{\pi}{6}) = \frac{1}{2}$.
When $x = \frac{\pi}{4}$,$u = \sin(\frac{\pi}{2}) = 1$.
$I = \int_{1/2}^{1} u^3 \frac{du}{2} = \frac{1}{2} [\frac{u^4}{4}]_{1/2}^{1} = \frac{1}{8} [1^4 - (\frac{1}{2})^4] = \frac{1}{8} [1 - \frac{1}{16}] = \frac{1}{8} \times \frac{15}{16} = \frac{15}{128}$.

(A) Let $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.
Substitute $x-1 = \sqrt{3} \tan \theta$,so $dx = \sqrt{3} \sec^2 \theta \, d\theta$.
When $x=1$,$\tan \theta = 0 \implies \theta = 0$.
When $x=2$,$\tan \theta = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6}$.
$I = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta \, d\theta}{(\sqrt{3} \sec \theta)^3} = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta}{3\sqrt{3} \sec^3 \theta} \, d\theta$.
$I = \frac{1}{3} \int_{0}^{\pi/6} \cos \theta \, d\theta = \frac{1}{3} [\sin \theta]_{0}^{\pi/6} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
Given $\frac{k}{k+5} = \frac{1}{6}$,we have $6k = k+5$,which implies $5k = 5$,so $k = 1$.

(A) Let $I = \int_{7\pi/4}^{7\pi/3} \sqrt{\tan^2 x} \, dx$.
Since $\sqrt{\tan^2 x} = |\tan x|$,we evaluate the integral in the interval $[7\pi/4, 7\pi/3]$.
In the interval $[7\pi/4, 2\pi]$,$\tan x$ is negative,so $|\tan x| = -\tan x$.
In the interval $[2\pi, 7\pi/3]$,$\tan x$ is positive,so $|\tan x| = \tan x$.
Thus,$I = \int_{7\pi/4}^{2\pi} -\tan x \, dx + \int_{2\pi}^{7\pi/3} \tan x \, dx$.
$I = [\log|\cos x|]_{7\pi/4}^{2\pi} + [-\log|\cos x|]_{2\pi}^{7\pi/3}$.
$I = (\log|\cos 2\pi| - \log|\cos(7\pi/4)|) - (\log|\cos(7\pi/3)| - \log|\cos 2\pi|)$.
Since $\cos 2\pi = 1$,$\log 1 = 0$.
$I = -\log|\cos(7\pi/4)| - \log|\cos(7\pi/3)| = -\log(1/\sqrt{2}) - \log(1/2)$.
$I = \log(\sqrt{2}) + \log(2) = \log(\sqrt{2} \times 2) = \log(2\sqrt{2})$.

(D) Let $I = \int_{0}^{0.9} [x - 2[x]] \, dx$.
Since $0 \le x < 0.9$,the greatest integer function $[x] = 0$.
Substituting $[x] = 0$ into the integral,we get:
$I = \int_{0}^{0.9} [x - 2(0)] \, dx = \int_{0}^{0.9} [x] \, dx$.
Since $0 \le x < 0.9$,$[x] = 0$ for all $x$ in the interval $[0, 0.9)$.
Therefore,$I = \int_{0}^{0.9} 0 \, dx = 0$.

(D) Let $f(x) = x^4 - 2x^2$. To minimize the integral $I = \int_a^b f(x) dx$,we need to integrate over the region where $f(x)$ is negative.
$f(x) = x^2(x^2 - 2) = x^2(x - \sqrt{2})(x + \sqrt{2})$.
The function $f(x)$ is negative in the interval $(-\sqrt{2}, \sqrt{2})$.
Since the integral of a negative function over its entire negative region yields the minimum possible value,we set $a = -\sqrt{2}$ and $b = \sqrt{2}$.
Thus,the ordered pair $(a, b)$ is $(-\sqrt{2}, \sqrt{2})$.

(B) Let $I = \int_{\pi /6}^{\pi /4} \frac{dx}{\sin 2x (\tan^5 x + \cot^5 x)}$.
Using $\sin 2x = \frac{2\tan x}{1+\tan^2 x}$,we have $\frac{1}{\sin 2x} = \frac{1+\tan^2 x}{2\tan x}$.
$I = \int_{\pi /6}^{\pi /4} \frac{(1+\tan^2 x) dx}{2\tan x (\tan^5 x + \cot^5 x)}$.
Let $t = \tan x$,then $dt = \sec^2 x dx = (1+\tan^2 x) dx$.
When $x = \pi/6, t = 1/\sqrt{3}$. When $x = \pi/4, t = 1$.
$I = \int_{1/\sqrt{3}}^{1} \frac{dt}{2t(t^5 + 1/t^5)} = \int_{1/\sqrt{3}}^{1} \frac{t^4 dt}{2(t^{10} + 1)}$.
Let $u = t^5$,then $du = 5t^4 dt$,so $t^4 dt = du/5$.
When $t = 1/\sqrt{3}, u = (1/\sqrt{3})^5 = 1/(9\sqrt{3})$. When $t = 1, u = 1$.
$I = \frac{1}{2} \int_{1/(9\sqrt{3})}^{1} \frac{du/5}{u^2 + 1} = \frac{1}{10} [\tan^{-1} u]_{1/(9\sqrt{3})}^{1}$.
$I = \frac{1}{10} (\tan^{-1}(1) - \tan^{-1}(1/(9\sqrt{3}))) = \frac{1}{10} (\frac{\pi}{4} - \tan^{-1}(\frac{1}{9\sqrt{3}}))$.

(D) Let $I = \int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$.
Split the integral: $I = \int_{1}^{e} \left( \frac{x}{e} \right)^{2x} \log_{e} x \, dx - \int_{1}^{e} \left( \frac{e}{x} \right)^{x} \log_{e} x \, dx$.
For the first integral,let $u = \left( \frac{x}{e} \right)^{2x}$. Then $\log_{e} u = 2x \log_{e} \left( \frac{x}{e} \right) = 2x (\log_{e} x - 1)$.
Differentiating both sides: $\frac{1}{u} du = 2(\log_{e} x - 1) dx + 2x \cdot \frac{1}{x} dx = 2 \log_{e} x \, dx$.
So,$\log_{e} x \, dx = \frac{1}{2} \frac{du}{u}$.
When $x=1, u = (1/e)^2 = e^{-2}$. When $x=e, u = (e/e)^{2e} = 1$.
Thus,$\int_{1}^{e} \left( \frac{x}{e} \right)^{2x} \log_{e} x \, dx = \int_{e^{-2}}^{1} u \cdot \frac{1}{2} \frac{du}{u} = \frac{1}{2} [u]_{e^{-2}}^{1} = \frac{1}{2} (1 - e^{-2})$.
For the second integral,let $v = \left( \frac{e}{x} \right)^{x}$. Then $\log_{e} v = x \log_{e} \left( \frac{e}{x} \right) = x (1 - \log_{e} x)$.
Differentiating: $\frac{1}{v} dv = (1 - \log_{e} x) dx + x (-1/x) dx = -\log_{e} x \, dx$.
So,$\log_{e} x \, dx = -\frac{dv}{v}$.
When $x=1, v = (e/1)^1 = e$. When $x=e, v = (e/e)^e = 1$.
Thus,$\int_{1}^{e} \left( \frac{e}{x} \right)^{x} \log_{e} x \, dx = \int_{e}^{1} v \cdot (-dv/v) = -\int_{e}^{1} dv = [v]_{1}^{e} = e - 1$.
Combining these: $I = \frac{1}{2} (1 - e^{-2}) - (e - 1) = \frac{1}{2} - \frac{1}{2e^2} - e + 1 = \frac{3}{2} - e - \frac{1}{2e^2}$.

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x + \csc x} dx$.
Simplify the integrand: $\frac{\cot x}{\cot x + \csc x} = \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} = \frac{\cos x}{\cos x + 1}$.
Using the identity $\cos x = 2 \cos^2 \frac{x}{2} - 1$,we get $\frac{\cos x}{\cos x + 1} = \frac{2 \cos^2 \frac{x}{2} - 1}{2 \cos^2 \frac{x}{2}} = 1 - \frac{1}{2} \sec^2 \frac{x}{2}$.
Now integrate: $I = \int_0^{\frac{\pi}{2}} (1 - \frac{1}{2} \sec^2 \frac{x}{2}) dx$.
$I = [x - \tan \frac{x}{2}]_0^{\frac{\pi}{2}}$.
Evaluating at the limits: $I = (\frac{\pi}{2} - \tan \frac{\pi}{4}) - (0 - \tan 0) = \frac{\pi}{2} - 1$.
We can write this as $I = \frac{1}{2}(\pi - 2) = \frac{1}{2}(\pi + (-2))$.
Comparing with $m(\pi + n)$,we get $m = \frac{1}{2}$ and $n = -2$.
Therefore,$m \cdot n = \frac{1}{2} \times (-2) = -1$.

(B) Given the integral $\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\frac{9}{8}\right)$.
Using partial fractions,$\frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1}$.
Integrating both terms:
$\int_{\alpha}^{\alpha+1} \left( \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1} \right) dx = \left[ \log_{e}|x+\alpha| - \log_{e}|x+\alpha+1| \right]_{\alpha}^{\alpha+1} = \left[ \log_{e}\left| \frac{x+\alpha}{x+\alpha+1} \right| \right]_{\alpha}^{\alpha+1}$.
Evaluating at the limits:
$\log_{e}\left( \frac{2\alpha+1}{2\alpha+2} \right) - \log_{e}\left( \frac{2\alpha}{2\alpha+1} \right) = \log_{e}\left( \frac{(2\alpha+1)^2}{2\alpha(2\alpha+2)} \right) = \log_{e}\left( \frac{9}{8} \right)$.
Equating the arguments:
$\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)} = \frac{9}{8} \Rightarrow 8(4\alpha^2 + 4\alpha + 1) = 9(4\alpha^2 + 4\alpha)$.
$32\alpha^2 + 32\alpha + 8 = 36\alpha^2 + 36\alpha \Rightarrow 4\alpha^2 + 4\alpha - 8 = 0$.
$\alpha^2 + \alpha - 2 = 0 \Rightarrow (\alpha+2)(\alpha-1) = 0$.
Thus,$\alpha = 1$ or $\alpha = -2$. Comparing with the options,the correct answer is $-2$.

(A) The integral of $\frac{1}{x}$ with respect to $x$ is $\log_{e} |x|$.
Applying the fundamental theorem of calculus for the definite integral $\int_{a}^{b} \frac{1}{x} dx$:
$= [\log_{e} |x|]_{a}^{b}$
$= \log_{e} |b| - \log_{e} |a|$
Using the logarithmic property $\log_{e} m - \log_{e} n = \log_{e} \left( \frac{m}{n} \right)$:
$= \log_{e} \left( \frac{b}{a} \right)$.

(D) Given the equation: $2 \cot^{2} \theta - \frac{5}{\sin \theta} + 4 = 0$.
Since $\cot^{2} \theta = \frac{\cos^{2} \theta}{\sin^{2} \theta} = \frac{1 - \sin^{2} \theta}{\sin^{2} \theta}$,we substitute this into the equation:
$2 \left( \frac{1 - \sin^{2} \theta}{\sin^{2} \theta} \right) - \frac{5}{\sin \theta} + 4 = 0$.
Multiplying by $\sin^{2} \theta$ (where $\sin \theta \neq 0$):
$2(1 - \sin^{2} \theta) - 5 \sin \theta + 4 \sin^{2} \theta = 0$.
$2 - 2 \sin^{2} \theta - 5 \sin \theta + 4 \sin^{2} \theta = 0$.
$2 \sin^{2} \theta - 5 \sin \theta + 2 = 0$.
Factoring the quadratic equation:
$(2 \sin \theta - 1)(\sin \theta - 2) = 0$.
Thus,$\sin \theta = \frac{1}{2}$ or $\sin \theta = 2$. Since $\sin \theta$ cannot be $2$,we have $\sin \theta = \frac{1}{2}$.
In the interval $(0, 2\pi) - \{\pi\}$,the solutions are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
Here,$\theta_{1} = \frac{\pi}{6}$ and $\theta_{2} = \frac{5\pi}{6}$.
Now,calculate the integral:
$I = \int_{\pi/6}^{5\pi/6} \cos^{2} 3\theta \, d\theta = \int_{\pi/6}^{5\pi/6} \frac{1 + \cos 6\theta}{2} \, d\theta$.
$I = \frac{1}{2} \left[ \theta + \frac{\sin 6\theta}{6} \right]_{\pi/6}^{5\pi/6}$.
$I = \frac{1}{2} \left[ \left( \frac{5\pi}{6} + \frac{\sin 5\pi}{6} \right) - \left( \frac{\pi}{6} + \frac{\sin \pi}{6} \right) \right] = \frac{1}{2} \left( \frac{5\pi}{6} - \frac{\pi}{6} \right) = \frac{1}{2} \left( \frac{4\pi}{6} \right) = \frac{\pi}{3}$.

(C) Given the integral $I = \int_{-1}^{2} e^{-\alpha |x|} dx$.
Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$,we split the integral:
$I = \int_{-1}^{0} e^{\alpha x} dx + \int_{0}^{2} e^{-\alpha x} dx$.
Evaluating the integrals:
$I = \left[ \frac{e^{\alpha x}}{\alpha} \right]_{-1}^{0} + \left[ \frac{e^{-\alpha x}}{-\alpha} \right]_{0}^{2} = \left( \frac{1}{\alpha} - \frac{e^{-\alpha}}{\alpha} \right) + \left( \frac{1}{\alpha} - \frac{e^{-2\alpha}}{\alpha} \right) = \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha}$.
Now,substitute this into the given equation $4\alpha I = 5$:
$4\alpha \left( \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha} \right) = 5$.
$4(2 - e^{-\alpha} - e^{-2\alpha}) = 5 \Rightarrow 8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5$.
$4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$.
Let $t = e^{-\alpha}$. Then $4t^2 + 4t - 3 = 0$.
Solving the quadratic equation: $t = \frac{-4 \pm \sqrt{16 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8}$.
Since $t = e^{-\alpha} > 0$,we take $t = \frac{4}{8} = \frac{1}{2}$.
$e^{-\alpha} = \frac{1}{2} \Rightarrow e^{\alpha} = 2 \Rightarrow \alpha = \log_{e} 2$.

(A) Let $f(x) = \frac{1}{\sqrt{2x^{3}-9x^{2}+12x+4}}$.
To find the range of $I$,we examine the behavior of $g(x) = 2x^{3}-9x^{2}+12x+4$ on the interval $[1, 2]$.
$g'(x) = 6x^{2}-18x+12 = 6(x^{2}-3x+2) = 6(x-1)(x-2)$.
Since $g'(x) \leq 0$ for $x \in [1, 2]$,$g(x)$ is a decreasing function on $[1, 2]$.
Thus,$g(2) \leq g(x) \leq g(1)$.
$g(1) = 2-9+12+4 = 9$.
$g(2) = 16-36+24+4 = 8$.
Therefore,$8 \leq g(x) \leq 9$ for $x \in [1, 2]$.
Taking the square root and reciprocal,we get $\frac{1}{3} \leq \frac{1}{\sqrt{g(x)}} \leq \frac{1}{\sqrt{8}}$.
Integrating from $1$ to $2$: $\int_{1}^{2} \frac{1}{3} dx < I < \int_{1}^{2} \frac{1}{\sqrt{8}} dx$.
$\frac{1}{3}(2-1) < I < \frac{1}{\sqrt{8}}(2-1)$.
$\frac{1}{3} < I < \frac{1}{\sqrt{8}}$.
Squaring the inequality: $\frac{1}{9} < I^{2} < \frac{1}{8}$.

(C) Given $f(x) = a + bx + cx^2$.
Calculating the definite integral:
$\int_{0}^{1} (a + bx + cx^2) dx = [ax + \frac{bx^2}{2} + \frac{cx^3}{3}]_{0}^{1} = a + \frac{b}{2} + \frac{c}{3}$.
Simplifying the expression:
$a + \frac{b}{2} + \frac{c}{3} = \frac{6a + 3b + 2c}{6}$.
Now,evaluate the terms in option $C$:
$f(0) = a$.
$f(1) = a + b + c$.
$f(\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}$.
Substituting these into $\frac{1}{6} \{f(0) + f(1) + 4f(\frac{1}{2})\}$:
$= \frac{1}{6} \{a + (a + b + c) + 4(a + \frac{b}{2} + \frac{c}{4})\}$
$= \frac{1}{6} \{a + a + b + c + 4a + 2b + c\}$
$= \frac{1}{6} \{6a + 3b + 2c\} = a + \frac{b}{2} + \frac{c}{3}$.
Thus,the correct option is $C$.

By definition,$\int_a^b f(x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
In this problem,$a = 0$,$b = 2$,$f(x) = x^2 + 1$,and $h = \frac{2-0}{n} = \frac{2}{n}$.
Therefore,$\int_0^2 (x^2 + 1) dx = 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(\frac{2r}{n})$.
$= 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} [(\frac{2r}{n})^2 + 1] = 2 \lim_{n \to \infty} \frac{1}{n} [\sum_{r=0}^{n-1} \frac{4r^2}{n^2} + \sum_{r=0}^{n-1} 1]$.
$= 2 \lim_{n \to \infty} [\frac{4}{n^3} \sum_{r=0}^{n-1} r^2 + \frac{1}{n} \sum_{r=0}^{n-1} 1]$.
Using $\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$ and $\sum_{r=0}^{n-1} 1 = n$,we get:
$= 2 \lim_{n \to \infty} [\frac{4}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} + \frac{1}{n} \cdot n]$.
$= 2 \lim_{n \to \infty} [\frac{2}{3} \cdot \frac{(n-1)(2n-1)}{n^2} + 1] = 2 [\frac{2}{3} \cdot 2 + 1] = 2 [\frac{4}{3} + 1] = 2 [\frac{7}{3}] = \frac{14}{3}$.

(D) Let $I = \int_{0}^{5}(x+1) d x$.
It is known that,$\int_a^b f (x)dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
Here,$a=0, b=5$,and $f(x)=(x+1)$.
$\Rightarrow h = \frac{5-0}{n} = \frac{5}{n}$.
$\therefore \int_0^5 (x + 1) dx = 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(r \cdot \frac{5}{n}\right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left( \frac{5r}{n} + 1 \right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ \sum_{r=0}^{n-1} 1 + \frac{5}{n} \sum_{r=0}^{n-1} r \right]$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{5}{n} \cdot \frac{(n-1)n}{2} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5(n-1)}{2n} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5}{2} \left( 1 - \frac{1}{n} \right) \right]$.
$= 5 \left[ 1 + \frac{5}{2} \right] = 5 \left[ \frac{7}{2} \right] = \frac{35}{2}$.

(N/A) Let $I = \int_{1}^{4}(x^{2}-x) dx$.
Using the property of linearity of integrals,we have:
$I = \int_{1}^{4} x^{2} dx - \int_{1}^{4} x dx = I_{1} - I_{2}$,where $I_{1} = \int_{1}^{4} x^{2} dx$ and $I_{2} = \int_{1}^{4} x dx$.
The definition of a definite integral as the limit of a sum is:
$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
For $I_{1} = \int_{1}^{4} x^{2} dx$:
$a = 1, b = 4, f(x) = x^{2}, h = \frac{4-1}{n} = \frac{3}{n}$.
$I_{1} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + r \cdot \frac{3}{n})^{2} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + \frac{6r}{n} + \frac{9r^{2}}{n^{2}})$.
$I_{1} = \lim_{n \to \infty} \frac{3}{n} [n + \frac{6}{n} \frac{(n-1)n}{2} + \frac{9}{n^{2}} \frac{(n-1)n(2n-1)}{6}] = 3 + 9 + \frac{27}{2} = 21$.
For $I_{2} = \int_{1}^{4} x dx$:
$a = 1, b = 4, f(x) = x, h = \frac{3}{n}$.
$I_{2} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + \frac{3r}{n}) = \lim_{n \to \infty} \frac{3}{n} [n + \frac{3}{n} \frac{(n-1)n}{2}] = 3 + \frac{9}{2} = \frac{15}{2}$.
Thus,$I = I_{1} - I_{2} = 21 - \frac{15}{2} = \frac{42-15}{2} = \frac{27}{2}$.

(N/A) Let $I = \int_{2}^{3} x^{2} dx$.
We know that the antiderivative of $x^{2}$ is $\int x^{2} dx = \frac{x^{3}}{3} = F(x)$.
By the Second Fundamental Theorem of Calculus,we have:
$I = F(3) - F(2)$
$I = \frac{3^{3}}{3} - \frac{2^{3}}{3}$
$I = \frac{27}{3} - \frac{8}{3}$
$I = \frac{19}{3}$

(B) Let $I = \int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$.
Using partial fractions,we decompose the integrand:
$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$
$x = A(x+2) + B(x+1)$
Setting $x = -1$,we get $A = -1$. Setting $x = -2$,we get $B = 2$.
So,$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$.
The indefinite integral is:
$\int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx = -\log |x+1| + 2 \log |x+2| = F(x)$.
Applying the Fundamental Theorem of Calculus:
$I = F(2) - F(1)$
$I = [-\log(3) + 2 \log(4)] - [-\log(2) + 2 \log(3)]$
$I = -\log(3) + 2 \log(4) + \log(2) - 2 \log(3)$
$I = 2 \log(4) + \log(2) - 3 \log(3)$
$I = \log(16) + \log(2) - \log(27)$
$I = \log \left( \frac{16 \times 2}{27} \right) = \log \left( \frac{32}{27} \right)$.

(C) Let $I = \int_{-1}^{1} (x+1) d x$.
First,find the indefinite integral:
$\int (x+1) d x = \frac{x^2}{2} + x = F(x)$.
By the second fundamental theorem of calculus,we have:
$I = F(1) - F(-1)$.
Substitute the limits:
$F(1) = \frac{1^2}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}$.
$F(-1) = \frac{(-1)^2}{2} + (-1) = \frac{1}{2} - 1 = -\frac{1}{2}$.
Calculate the difference:
$I = \frac{3}{2} - (-\frac{1}{2}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$.

(B) Let $I = \int_{2}^{3} \frac{1}{x} d x$.
We know that the antiderivative of $\frac{1}{x}$ is $\log |x|$.
By the second fundamental theorem of calculus,$\int_{a}^{b} f(x) d x = F(b) - F(a)$,where $F(x)$ is the antiderivative of $f(x)$.
Here,$F(x) = \log |x|$.
Therefore,$I = F(3) - F(2) = \log |3| - \log |2|$.
Using the logarithmic property $\log a - \log b = \log \frac{a}{b}$,we get $I = \log \frac{3}{2}$.

(D) Let $I = \int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) dx$.
First,find the indefinite integral:
$\int (4x^{3} - 5x^{2} + 6x + 9) dx = 4(\frac{x^{4}}{4}) - 5(\frac{x^{3}}{3}) + 6(\frac{x^{2}}{2}) + 9x = x^{4} - \frac{5x^{3}}{3} + 3x^{2} + 9x = F(x)$.
By the second fundamental theorem of calculus,$I = F(2) - F(1)$.
$F(2) = (2)^{4} - \frac{5(2)^{3}}{3} + 3(2)^{2} + 9(2) = 16 - \frac{40}{3} + 12 + 18 = 46 - \frac{40}{3} = \frac{138 - 40}{3} = \frac{98}{3}$.
$F(1) = (1)^{4} - \frac{5(1)^{3}}{3} + 3(1)^{2} + 9(1) = 1 - \frac{5}{3} + 3 + 9 = 13 - \frac{5}{3} = \frac{39 - 5}{3} = \frac{34}{3}$.
Therefore,$I = \frac{98}{3} - \frac{34}{3} = \frac{64}{3}$.

(A) Let $I = \int_{0}^{\frac{\pi}{4}} \sin 2x \,dx$.
The antiderivative of $\sin 2x$ is $\int \sin 2x \,dx = -\frac{\cos 2x}{2}$.
By the Second Fundamental Theorem of Calculus,we have:
$I = \left[ -\frac{\cos 2x}{2} \right]_{0}^{\frac{\pi}{4}}$
$I = -\frac{1}{2} \left[ \cos 2\left(\frac{\pi}{4}\right) - \cos 2(0) \right]$
$I = -\frac{1}{2} \left[ \cos\left(\frac{\pi}{2}\right) - \cos(0) \right]$
Since $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$,we get:
$I = -\frac{1}{2} [0 - 1] = \frac{1}{2}$.