Fundamental definite integration Questions in English

(B) We know that $1 + \cos 2x = 2 \cos^2 x$.
Substituting this into the integral,we get:
$I = \int_{0}^{\pi} \sqrt{\frac{2 \cos^2 x}{2}} \, dx = \int_{0}^{\pi} |\cos x| \, dx$.
Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$I = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{\pi} \cos x \, dx$.
Evaluating the integrals:
$I = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$.
$I = (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2))$.
$I = (1 - 0) - (0 - 1) = 1 + 1 = 2$.

(A) Let $I = \int_{0}^{9} [\sqrt{x} + 2] \,dx$. Since $2$ is an integer,we can write this as $I = \int_{0}^{9} ([\sqrt{x}] + 2) \,dx = \int_{0}^{9} [\sqrt{x}] \,dx + \int_{0}^{9} 2 \,dx$.
First,evaluate $\int_{0}^{9} [\sqrt{x}] \,dx$ by splitting the interval based on the values of $[\sqrt{x}]$:
For $0 \le x < 1$,$[\sqrt{x}] = 0$.
For $1 \le x < 4$,$[\sqrt{x}] = 1$.
For $4 \le x < 9$,$[\sqrt{x}] = 2$.
Thus,$\int_{0}^{9} [\sqrt{x}] \,dx = \int_{0}^{1} 0 \,dx + \int_{1}^{4} 1 \,dx + \int_{4}^{9} 2 \,dx = 0 + (4 - 1) + 2(9 - 4) = 3 + 10 = 13$.
Next,$\int_{0}^{9} 2 \,dx = [2x]_{0}^{9} = 18$.
Therefore,$I = 13 + 18 = 31$.

(C) Let $I = \int_{0}^{\sqrt{2}} [x^2] \, dx$.
Since the function $[x^2]$ changes its value at points where $x^2$ is an integer,we split the interval $[0, \sqrt{2}]$.
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x \le \sqrt{2}$,$1 \le x^2 \le 2$,so $[x^2] = 1$.
Thus,$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx$.
$I = 0 + [x]_{1}^{\sqrt{2}}$.
$I = \sqrt{2} - 1$.

(C) Given $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$.
Since the integrand involves an absolute value,we split the integral at $x = 1/2$:
$I = \int_{0}^{1/2} x \left( -\left( x - \frac{1}{2} \right) \right) dx + \int_{1/2}^{1} x \left( x - \frac{1}{2} \right) dx$.
$I = \int_{0}^{1/2} \left( \frac{1}{2}x - x^2 \right) dx + \int_{1/2}^{1} \left( x^2 - \frac{1}{2}x \right) dx$.
Evaluating the integrals:
$I = \left[ \frac{x^2}{4} - \frac{x^3}{3} \right]_{0}^{1/2} + \left[ \frac{x^3}{3} - \frac{x^2}{4} \right]_{1/2}^{1}$.
$I = \left( \frac{(1/2)^2}{4} - \frac{(1/2)^3}{3} \right) + \left( \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{(1/2)^3}{3} - \frac{(1/2)^2}{4} \right) \right)$.
$I = \left( \frac{1}{16} - \frac{1}{24} \right) + \left( \frac{1}{12} - \left( \frac{1}{24} - \frac{1}{16} \right) \right)$.
$I = \left( \frac{3-2}{48} \right) + \left( \frac{1}{12} - \left( \frac{2-3}{48} \right) \right) = \frac{1}{48} + \frac{1}{12} + \frac{1}{48} = \frac{2}{48} + \frac{4}{48} = \frac{6}{48} = \frac{1}{8}$.

(A) We need to evaluate the definite integral $I = \int_{0}^{8} |x - 5| \, dx$.
Since the integrand $|x - 5|$ changes its definition at $x = 5$,we split the integral at $x = 5$:
$I = \int_{0}^{5} -(x - 5) \, dx + \int_{5}^{8} (x - 5) \, dx$.
Evaluating the first part: $\int_{0}^{5} (5 - x) \, dx = [5x - \frac{x^2}{2}]_{0}^{5} = (25 - \frac{25}{2}) - 0 = \frac{25}{2} = 12.5$.
Evaluating the second part: $\int_{5}^{8} (x - 5) \, dx = [\frac{x^2}{2} - 5x]_{5}^{8} = (\frac{64}{2} - 40) - (\frac{25}{2} - 25) = (32 - 40) - (12.5 - 25) = -8 - (-12.5) = 4.5$.
Adding both parts: $I = 12.5 + 4.5 = 17$.

(D) Let $I = \int_{0}^{2} |x - 1| \, dx$.
Since the function $|x - 1|$ changes its sign at $x = 1$,we split the integral at $x = 1$:
$I = \int_{0}^{1} -(x - 1) \, dx + \int_{1}^{2} (x - 1) \, dx$.
Evaluating the first integral: $\int_{0}^{1} (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$.
Evaluating the second integral: $\int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} = (\frac{4}{2} - 2) - (\frac{1}{2} - 1) = (2 - 2) - (-\frac{1}{2}) = \frac{1}{2}$.
Adding both parts: $I = \frac{1}{2} + \frac{1}{2} = 1$.

(D) We evaluate the integral $\int_{-2}^{2} |[x]| \, dx$ by splitting the interval at the integers where the greatest integer function $[x]$ changes value:
$\int_{-2}^{2} |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^{0} |[x]| \, dx + \int_{0}^{1} |[x]| \, dx + \int_{1}^{2} |[x]| \, dx$
For $x \in [-2, -1)$,$[x] = -2$,so $|[x]| = 2$.
For $x \in [-1, 0)$,$[x] = -1$,so $|[x]| = 1$.
For $x \in [0, 1)$,$[x] = 0$,so $|[x]| = 0$.
For $x \in [1, 2)$,$[x] = 1$,so $|[x]| = 1$.
Substituting these values:
$= \int_{-2}^{-1} 2 \, dx + \int_{-1}^{0} 1 \, dx + \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx$
$= 2[x]_{-2}^{-1} + [x]_{-1}^{0} + 0 + [x]_{1}^{2}$
$= 2(-1 - (-2)) + (0 - (-1)) + 0 + (2 - 1)$
$= 2(1) + 1 + 1 = 4.$

(B) Given the integral $\int_{a}^{b} \frac{x}{|x|} dx$ with the condition $a < b < 0$.
Since $x < 0$ for all $x$ in the interval $[a, b]$,we have $|x| = -x$.
Therefore,the integrand becomes $\frac{x}{|x|} = \frac{x}{-x} = -1$.
Now,evaluating the definite integral:
$\int_{a}^{b} (-1) dx = -[x]_{a}^{b} = -(b - a) = a - b$.
Since $a < b < 0$,we know that $|a| = -a$ and $|b| = -b$.
Thus,$a - b = (-|a|) - (-|b|) = |b| - |a|$.
Hence,the correct option is $B$.

(D) We need to evaluate the integral $I = \int_{-2}^{3} |1 - x^2| dx$.
The expression $|1 - x^2|$ changes sign at $x = -1$ and $x = 1$.
For $x \in [-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.
For $x \in [-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$.
For $x \in [1, 3]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.
Thus,$I = \int_{-2}^{-1} (x^2 - 1) dx + \int_{-1}^{1} (1 - x^2) dx + \int_{1}^{3} (x^2 - 1) dx$.
Evaluating each integral:
$\int_{-2}^{-1} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{-1}^{1} (1 - x^2) dx = [x - \frac{x^3}{3}]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{1}^{3} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{3} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$.
Summing these values: $I = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$.

(A) Given $f(x) = |x - 1|$.
We know that $|x - 1| = (1 - x)$ for $x < 1$ and $|x - 1| = (x - 1)$ for $x \ge 1$.
Therefore,$\int_0^2 |x - 1| dx = \int_0^1 (1 - x) dx + \int_1^2 (x - 1) dx$.
Evaluating the first integral: $\int_0^1 (1 - x) dx = [x - \frac{x^2}{2}]_0^1 = (1 - \frac{1}{2}) - (0 - 0) = \frac{1}{2}$.
Evaluating the second integral: $\int_1^2 (x - 1) dx = [\frac{x^2}{2} - x]_1^2 = (\frac{4}{2} - 2) - (\frac{1}{2} - 1) = (2 - 2) - (-\frac{1}{2}) = \frac{1}{2}$.
Adding both parts: $\frac{1}{2} + \frac{1}{2} = 1$.

(C) Let $I = \int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx$.
Consider the substitution $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) \, dx$,which implies $-( \sin x - \cos x) \, dx = dt$.
Now,change the limits of integration:
When $x = 0$,$t = \sin(0) + \cos(0) = 0 + 1 = 1$.
When $x = \frac{\pi}{2}$,$t = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1$.
Substituting these into the integral,we get:
$I = \int_{1}^{1} - \log(t) \, dt$.
Since the lower limit and upper limit are the same,the value of the definite integral is $0$,because $\int_{a}^{a} f(t) \, dt = 0$.

(C) Given the function $L(x) = \int_1^x \frac{1}{t} dt$.
Evaluating the integral,we get $L(x) = [\ln |t|]_1^x$.
$L(x) = \ln |x| - \ln |1| = \ln |x|$.
Now,consider $L(xy) = \ln |xy| = \ln |x| + \ln |y|$.
Since $L(x) = \ln |x|$ and $L(y) = \ln |y|$,we have $L(xy) = L(x) + L(y)$.
Thus,the correct option is $C$.

(C) For $0 < x < 1$,we have $x^2 < x$,which implies $e^{x^2} < e^x$. However,a simpler approach is to observe the bounds of the function $f(x) = e^{x^2}$ on the interval $[0, 1]$.
Since $0 \le x^2 \le 1$ for $x \in [0, 1]$,it follows that $e^0 \le e^{x^2} \le e^1$,which means $1 \le e^{x^2} \le e$.
Integrating this inequality from $0$ to $1$ with respect to $x$:
$\int_0^1 1 dx < \int_0^1 e^{x^2} dx < \int_0^1 e dx$
$[x]_0^1 < \int_0^1 e^{x^2} dx < [ex]_0^1$
$1 < \int_0^1 e^{x^2} dx < e$.
Therefore,the value of the integral lies in the interval $(1, e)$.

(C) Given the equation: $\int_0^3 {(3ax^2 + 2bx + c)\,dx} = \int_1^3 {(3ax^2 + 2bx + c)\,dx}$.
Using the property of definite integrals $\int_0^3 f(x)dx = \int_0^1 f(x)dx + \int_1^3 f(x)dx$,we can rewrite the equation as:
$\int_0^1 {(3ax^2 + 2bx + c)\,dx} + \int_1^3 {(3ax^2 + 2bx + c)\,dx} = \int_1^3 {(3ax^2 + 2bx + c)\,dx}$.
Subtracting $\int_1^3 {(3ax^2 + 2bx + c)\,dx}$ from both sides,we get:
$\int_0^1 {(3ax^2 + 2bx + c)\,dx} = 0$.
Now,evaluating the integral:
$\left[ ax^3 + bx^2 + cx \right]_0^1 = 0$.
Substituting the limits:
$(a(1)^3 + b(1)^2 + c(1)) - (a(0)^3 + b(0)^2 + c(0)) = 0$.
$a + b + c = 0$.

(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$.
Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is even and $0$ if $f(x)$ is odd:
$I = \int_{-\pi}^{\pi} \cos^2 px dx + \int_{-\pi}^{\pi} \sin^2 qx dx - 2 \int_{-\pi}^{\pi} \sin qx \cos px dx$.
Since $\sin qx \cos px$ is an odd function,its integral over $[-\pi, \pi]$ is $0$.
$I = \int_{-\pi}^{\pi} \frac{1 + \cos 2px}{2} dx + \int_{-\pi}^{\pi} \frac{1 - \cos 2qx}{2} dx$.
$I = \frac{1}{2} [x + \frac{\sin 2px}{2p}]_{-\pi}^{\pi} + \frac{1}{2} [x - \frac{\sin 2qx}{2q}]_{-\pi}^{\pi}$.
$I = \frac{1}{2} [(\pi - (-\pi)) + 0] + \frac{1}{2} [(\pi - (-\pi)) - 0] = \frac{1}{2} (2\pi) + \frac{1}{2} (2\pi) = \pi + \pi = 2\pi$.

(D) We are given the integral $I = \int_0^1 (1 + e^{-x^2}) \,dx$.
By the linearity property of definite integrals,we can split this into two parts:
$I = \int_0^1 1 \,dx + \int_0^1 e^{-x^2} \,dx$.
The first part is $\int_0^1 1 \,dx = [x]_0^1 = 1 - 0 = 1$.
The second part is $\int_0^1 e^{-x^2} \,dx$.
The function $e^{-x^2}$ does not have an elementary antiderivative. Therefore,the integral $\int_0^1 e^{-x^2} \,dx$ cannot be expressed in terms of simple algebraic or transcendental functions.
Thus,the value of the integral is $1 + \int_0^1 e^{-x^2} \,dx$.
Comparing this with the given options,none of the options $A, B,$ or $C$ match this result.
Therefore,the correct option is $D$.

(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$.
Since $f(x) = \sin^2 x$ is an even function,we have $I = 2 \int_0^{\pi/2} \sin^2 x \, dx$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get:
$I = 2 \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx$
$I = \int_0^{\pi/2} (1 - \cos 2x) \, dx$
$I = [x - \frac{\sin 2x}{2}]_0^{\pi/2}$
$I = (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2})$
$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

(A) To evaluate $I = \int_0^{\pi /2} \cos^2 x \, dx$,we use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.
Substituting this into the integral:
$I = \int_0^{\pi /2} \frac{1 + \cos(2x)}{2} \, dx$
$I = \frac{1}{2} \int_0^{\pi /2} (1 + \cos(2x)) \, dx$
$I = \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_0^{\pi /2}$
Evaluating at the limits:
$I = \frac{1}{2} \left[ (\frac{\pi}{2} + \frac{\sin(\pi)}{2}) - (0 + \frac{\sin(0)}{2}) \right]$
Since $\sin(\pi) = 0$ and $\sin(0) = 0$:
$I = \frac{1}{2} [\frac{\pi}{2} + 0 - 0] = \frac{\pi}{4}$.

(D) Let $I = \int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)}$.
Using partial fractions,we can write $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}$.
Solving for $A$ and $B$,we get $x^2 = A(x^2 + b^2) + B(x^2 + a^2)$.
Setting $x^2 = -a^2$,we get $-a^2 = A(b^2 - a^2)$,so $A = \frac{a^2}{a^2 - b^2}$.
Setting $x^2 = -b^2$,we get $-b^2 = B(a^2 - b^2)$,so $B = \frac{-b^2}{a^2 - b^2}$.
Thus,$I = \frac{1}{a^2 - b^2} \int_0^\infty \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right) dx$.
$I = \frac{1}{a^2 - b^2} \left[ a^2 \cdot \frac{1}{a} \tan^{-1}(\frac{x}{a}) - b^2 \cdot \frac{1}{b} \tan^{-1}(\frac{x}{b}) \right]_0^\infty$.
$I = \frac{1}{a^2 - b^2} \left[ a \cdot \frac{\pi}{2} - b \cdot \frac{\pi}{2} \right] = \frac{\pi}{2(a^2 - b^2)} (a - b) = \frac{\pi}{2(a + b)}$.

(B) Let $I = \int_0^\infty \frac{x^3 \, dx}{(x^2 + 4)^2}$.
Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.
When $x = 0$,$t = 0$. When $x \to \infty$,$t \to \infty$.
$I = \int_0^\infty \frac{t \cdot (x \, dx)}{(t + 4)^2} = \int_0^\infty \frac{t \cdot \frac{dt}{2}}{(t + 4)^2} = \frac{1}{2} \int_0^\infty \frac{t}{(t + 4)^2} \, dt$.
We can rewrite the integrand as $\frac{t}{(t + 4)^2} = \frac{t + 4 - 4}{(t + 4)^2} = \frac{1}{t + 4} - \frac{4}{(t + 4)^2}$.
$I = \frac{1}{2} \left[ \int_0^\infty \frac{1}{t + 4} \, dt - \int_0^\infty \frac{4}{(t + 4)^2} \, dt \right]$.
$I = \frac{1}{2} \left[ \ln(t + 4) + \frac{4}{t + 4} \right]_0^\infty$.
As $t \to \infty$,$\ln(t + 4) \to \infty$ and $\frac{4}{t + 4} \to 0$.
Thus,the integral diverges to $\infty$.

(A) Let $I = \int_0^\infty \frac{x}{(1 + x)(1 + x^2)} dx$.
Using partial fractions,we write $\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$.
Solving for constants,we get $x = A(1 + x^2) + (Bx + C)(1 + x)$.
For $x = -1$,$-1 = A(2) \implies A = -1/2$.
Comparing coefficients of $x^2$,$0 = A + B \implies B = 1/2$.
Comparing constants,$0 = A + C \implies C = 1/2$.
Thus,$\frac{x}{(1 + x)(1 + x^2)} = \frac{1}{2} \left( \frac{x + 1}{1 + x^2} - \frac{1}{1 + x} \right) = \frac{1}{2} \left( \frac{x}{1 + x^2} + \frac{1}{1 + x^2} - \frac{1}{1 + x} \right)$.
Integrating from $0$ to $\infty$:
$I = \frac{1}{2} \left[ \frac{1}{2} \ln(1 + x^2) + \tan^{-1}(x) - \ln(1 + x) \right]_0^\infty$.
$I = \frac{1}{2} \left[ \frac{1}{2} \ln\left(\frac{1 + x^2}{(1 + x)^2}\right) + \tan^{-1}(x) \right]_0^\infty$.
As $x \to \infty$,$\frac{1 + x^2}{(1 + x)^2} \to 1$,so $\ln(1) = 0$.
$I = \frac{1}{2} [0 + \frac{\pi}{2} - (0 + 0)] = \frac{\pi}{4}$.

(B) Given $F(x) = \int_{5\pi /4}^x (3\sin u + 4\cos u) du$.
By the Fundamental Theorem of Calculus,$F'(x) = 3\sin x + 4\cos x$.
In the interval $\left[ \frac{5\pi}{4}, \frac{4\pi}{3} \right]$,both $\sin x$ and $\cos x$ are negative.
Specifically,$3\sin x + 4\cos x = 5 \left( \frac{3}{5}\sin x + \frac{4}{5}\cos x \right) = 5\sin(x + \alpha)$,where $\tan \alpha = 4/3$.
Since $F'(x) < 0$ for all $x \in \left[ \frac{5\pi}{4}, \frac{4\pi}{3} \right]$,the function $F(x)$ is strictly decreasing on this interval.
Therefore,the least value occurs at the right endpoint,$x = \frac{4\pi}{3}$.
$F\left( \frac{4\pi}{3} \right) = \int_{5\pi /4}^{4\pi /3} (3\sin u + 4\cos u) du = [-3\cos u + 4\sin u]_{5\pi /4}^{4\pi /3}$.
$= \left( -3\cos\frac{4\pi}{3} + 4\sin\frac{4\pi}{3} \right) - \left( -3\cos\frac{5\pi}{4} + 4\sin\frac{5\pi}{4} \right)$.
$= \left( -3(-\frac{1}{2}) + 4(-\frac{\sqrt{3}}{2}) \right) - \left( -3(-\frac{1}{\sqrt{2}}) + 4(-\frac{1}{\sqrt{2}}) \right)$.
$= (\frac{3}{2} - 2\sqrt{3}) - (\frac{3}{\sqrt{2}} - \frac{4}{\sqrt{2}}) = \frac{3}{2} - 2\sqrt{3} - (-\frac{1}{\sqrt{2}}) = \frac{3}{2} - 2\sqrt{3} + \frac{1}{\sqrt{2}}$.

(C) Let $I = \int_0^\infty e^{-2x} (\sin 2x + \cos 2x) \, dx$.
Using the standard integral formula $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx)$ and $\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$.
Here $a = -2$ and $b = 2$.
$\int_0^\infty e^{-2x} \sin 2x \, dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \sin 2x - 2 \cos 2x) \right]_0^\infty = \left[ \frac{e^{-2x}}{8} (-2 \sin 2x - 2 \cos 2x) \right]_0^\infty = 0 - (\frac{1}{8}(-2)) = \frac{1}{4}$.
$\int_0^\infty e^{-2x} \cos 2x \, dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \cos 2x + 2 \sin 2x) \right]_0^\infty = \left[ \frac{e^{-2x}}{8} (-2 \cos 2x + 2 \sin 2x) \right]_0^\infty = 0 - (\frac{1}{8}(-2)) = \frac{1}{4}$.
Adding both results: $I = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

(B) To evaluate the integral $\int_0^{b - c} f''(x + a) \, dx$,we use the Fundamental Theorem of Calculus.
Let $u = x + a$,then $du = dx$.
When $x = 0$,$u = a$.
When $x = b - c$,$u = b - c + a$.
The integral becomes:
$\int_a^{b - c + a} f''(u) \, du$
Since the antiderivative of $f''(u)$ is $f'(u)$,we have:
$[f'(u)]_a^{b - c + a} = f'(b - c + a) - f'(a)$.

(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.
By the Fundamental Theorem of Calculus,$F'(x) = |x|$.
Since $|x| \ge 0$ for all $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$,the function $F(x)$ is non-decreasing on the given interval.
Therefore,the maximum value occurs at the right endpoint $x = \frac{1}{2}$.
$F\left( \frac{1}{2} \right) = \int_1^{1/2} {|t| \, dt} = -\int_{1/2}^1 {|t| \, dt}$.
Since $t > 0$ in the interval $\left[ \frac{1}{2}, 1 \right]$,we have $|t| = t$.
$F\left( \frac{1}{2} \right) = -\int_{1/2}^1 {t \, dt} = -\left[ \frac{t^2}{2} \right]_{1/2}^1 = -\left( \frac{1}{2} - \frac{1}{8} \right) = -\left( \frac{4-1}{8} \right) = -\frac{3}{8}$.

(A) Let $I = \int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3}$.
Substitute $x = \tan \theta$,so $dx = \sec^2 \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x \to \infty$,$\theta \to \frac{\pi}{2}$.
The integral becomes:
$I = \int_0^{\pi/2} \frac{\sec^2 \theta \, d\theta}{(\tan \theta + \sec \theta)^3}$
$I = \int_0^{\pi/2} \frac{\sec^2 \theta \, d\theta}{(\frac{\sin \theta + 1}{\cos \theta})^3} = \int_0^{\pi/2} \frac{\sec^2 \theta \cos^3 \theta}{(1 + \sin \theta)^3} \, d\theta$
$I = \int_0^{\pi/2} \frac{\cos \theta}{(1 + \sin \theta)^3} \, d\theta$
Let $u = 1 + \sin \theta$,then $du = \cos \theta \, d\theta$.
When $\theta = 0, u = 1$. When $\theta = \frac{\pi}{2}, u = 2$.
$I = \int_1^2 u^{-3} \, du = \left[ \frac{u^{-2}}{-2} \right]_1^2 = \left[ -\frac{1}{2u^2} \right]_1^2$
$I = -\frac{1}{2(4)} - (-\frac{1}{2(1)}) = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}$.

(C) Let $I = \int_{0}^{\pi /2} (\sqrt{\sin \theta} \cos \theta)^3 d\theta = \int_{0}^{\pi /2} \sin^{3/2} \theta \cos^3 \theta d\theta$.
Using the Beta function formula $\int_{0}^{\pi /2} \sin^m \theta \cos^n \theta d\theta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$,where $m = 3/2$ and $n = 3$:
$I = \frac{\Gamma(\frac{3/2 + 1}{2}) \Gamma(\frac{3 + 1}{2})}{2 \Gamma(\frac{3/2 + 3 + 2}{2})} = \frac{\Gamma(5/4) \Gamma(2)}{2 \Gamma(13/4)}$.
Since $\Gamma(2) = 1! = 1$ and $\Gamma(13/4) = \Gamma(1 + 9/4) = \frac{9}{4} \Gamma(9/4) = \frac{9}{4} \cdot \frac{5}{4} \Gamma(5/4)$:
$I = \frac{\Gamma(5/4) \cdot 1}{2 \cdot (\frac{9}{4} \cdot \frac{5}{4} \Gamma(5/4))} = \frac{1}{2 \cdot \frac{45}{16}} = \frac{16}{90} = \frac{8}{45}$.

(D) Given $f(t) = \int_{-t}^{t} \frac{dx}{1 + x^2}$.
We know that $\int \frac{dx}{1 + x^2} = \tan^{-1}(x)$.
Applying the limits:
$f(t) = [\tan^{-1}(x)]_{-t}^{t} = \tan^{-1}(t) - \tan^{-1}(-t)$.
Since $\tan^{-1}(-t) = -\tan^{-1}(t)$,we have:
$f(t) = \tan^{-1}(t) - (-\tan^{-1}(t)) = 2\tan^{-1}(t)$.
Differentiating $f(t)$ with respect to $t$ using the chain rule:
$f'(t) = 2 \times \frac{1}{1 + t^2} = \frac{2}{1 + t^2}$.
Now,substitute $t = 1$:
$f'(1) = \frac{2}{1 + (1)^2} = \frac{2}{2} = 1$.

(C) Let $I = \int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$.
By the Fundamental Theorem of Calculus,$\int_{a}^{b} f'(x) dx = f(b) - f(a)$.
Here,$f(x) = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$.
So,$I = \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right]_{0}^{1}$.
Evaluating at the limits:
$I = \sin^{-1} \left( \frac{2(1)}{1+(1)^2} \right) - \sin^{-1} \left( \frac{2(0)}{1+(0)^2} \right)$.
$I = \sin^{-1} \left( \frac{2}{2} \right) - \sin^{-1} (0)$.
$I = \sin^{-1} (1) - \sin^{-1} (0)$.
Since $\sin^{-1} (1) = \frac{\pi}{2}$ and $\sin^{-1} (0) = 0$,
$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

(C) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$.
Using partial fractions,we write:
$\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$.
Solving for constants,we get $A = -1/2$,$B = 1/2$,and $C = 1/2$.
Thus,$I = \int_{0}^{\infty} \left( \frac{-1/2}{1 + x} + \frac{1/2 x + 1/2}{1 + x^2} \right) \, dx$.
$I = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{1 + x^2} \, dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x^2}$.
Evaluating the integral:
$I = \left[ -\frac{1}{2} \ln(1 + x) + \frac{1}{4} \ln(1 + x^2) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
$I = \left[ \frac{1}{4} \ln\left( \frac{1 + x^2}{(1 + x)^2} \right) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
As $x \to \infty$,$\frac{1 + x^2}{(1 + x)^2} \to 1$,so $\ln(1) = 0$.
$I = (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (0 + 0) = \frac{\pi}{4}$.

(B) Let $I = \int_0^a x^4 \sqrt{a^2 - x^2} \,dx$.
Put $x = a \sin \theta$,then $dx = a \cos \theta \,d\theta$.
When $x = 0$,$\theta = 0$. When $x = a$,$\theta = \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int_0^{\pi/2} (a \sin \theta)^4 \sqrt{a^2 - a^2 \sin^2 \theta} \cdot (a \cos \theta) \,d\theta$
$I = \int_0^{\pi/2} a^4 \sin^4 \theta \cdot (a \cos \theta) \cdot (a \cos \theta) \,d\theta$
$I = a^6 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \,d\theta$.
Using the reduction formula $\int_0^{\pi/2} \sin^m \theta \cos^n \theta \,d\theta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$:
$I = a^6 \cdot \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(\frac{4+2+2}{2})} = a^6 \cdot \frac{(\frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi}) \cdot (\frac{1}{2} \cdot \sqrt{\pi})}{2 \cdot \Gamma(4)}$
$I = a^6 \cdot \frac{\frac{3}{8} \pi}{2 \cdot 6} = a^6 \cdot \frac{3\pi}{96} = \frac{\pi}{32} a^6$.

(A) Let $I = \int_0^a x^2 (a^2 - x^2)^{3/2} dx$.
Substitute $x = a \sin \theta$,then $dx = a \cos \theta d\theta$.
When $x = 0$,$\theta = 0$. When $x = a$,$\theta = \frac{\pi}{2}$.
$I = \int_0^{\pi/2} (a^2 \sin^2 \theta) (a^2 - a^2 \sin^2 \theta)^{3/2} (a \cos \theta) d\theta$
$I = \int_0^{\pi/2} (a^2 \sin^2 \theta) (a^2 \cos^2 \theta)^{3/2} (a \cos \theta) d\theta$
$I = \int_0^{\pi/2} (a^2 \sin^2 \theta) (a^3 \cos^3 \theta) (a \cos \theta) d\theta$
$I = a^6 \int_0^{\pi/2} \sin^2 \theta \cos^4 \theta d\theta$
Using the formula $\int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$:
$I = a^6 \frac{\Gamma(\frac{3}{2}) \Gamma(\frac{5}{2})}{2 \Gamma(\frac{8}{2})} = a^6 \frac{(\frac{1}{2} \sqrt{\pi}) (\frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi})}{2 \cdot 3!} = a^6 \frac{\frac{3}{8} \pi}{2 \cdot 6} = a^6 \frac{3\pi}{96} = \frac{\pi a^6}{32}$.

(B) The given expression is $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{r}{{{n^2}}}{{\sec }^2}\frac{{{r^2}}}{{{n^2}}}} $.
This can be rewritten as $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {\frac{r}{n}{{\sec }^2}\left( {\frac{r}{n}} \right)^2} $.
Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} \right)} = \int_0^1 {f(x)dx} $.
Here,$f(x) = x \sec^2(x^2)$.
So,the limit is equal to $\int_0^1 {x{{\sec }^2}{x^2}dx} $.
Let $t = x^2$,then $dt = 2x dx$,or $x dx = \frac{1}{2} dt$.
When $x = 0$,$t = 0$. When $x = 1$,$t = 1$.
The integral becomes $\frac{1}{2} \int_0^1 {\sec^2 t dt} = \frac{1}{2} [\tan t]_0^1 = \frac{1}{2} \tan 1$.

(A) Let $f(x) = \frac{x}{x^3 + 16}$.
To find the range of $f(x)$ on $[0, 1]$,we check its derivative:
$f'(x) = \frac{(x^3 + 16)(1) - x(3x^2)}{(x^3 + 16)^2} = \frac{16 - 2x^3}{(x^3 + 16)^2}$.
Since $f'(x) > 0$ for all $x \in [0, 1]$,the function $f(x)$ is strictly increasing on $[0, 1]$.
Thus,the minimum value is $f(0) = 0$ and the maximum value is $f(1) = \frac{1}{1+16} = \frac{1}{17}$.
Using the property of definite integrals,$m(b-a) \le \int_a^b f(x) \, dx \le M(b-a)$,where $m$ and $M$ are the minimum and maximum values of $f(x)$ on $[a, b]$:
$0(1-0) \le \int_0^1 \frac{x}{x^3 + 16} \, dx \le \frac{1}{17}(1-0)$.
Therefore,the integral lies in the interval $[0, \frac{1}{17}]$.

(B) Given $g(2) = \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt$.
For $t \in [0, 1]$,we have $\frac{1}{2} \le f(t) \le 1$. Integrating this over $[0, 1]$:
$\int_0^1 \frac{1}{2} \, dt \le \int_0^1 f(t) \, dt \le \int_0^1 1 \, dt$
$\frac{1}{2} \le \int_0^1 f(t) \, dt \le 1 \quad \dots (i)$
For $t \in (1, 2]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating this over $(1, 2]$:
$\int_1^2 0 \, dt \le \int_1^2 f(t) \, dt \le \int_1^2 \frac{1}{2} \, dt$
$0 \le \int_1^2 f(t) \, dt \le \frac{1}{2} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$\frac{1}{2} + 0 \le \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt \le 1 + \frac{1}{2}$
$\frac{1}{2} \le g(2) \le \frac{3}{2}$.
Since $\frac{1}{2} \le g(2) \le \frac{3}{2}$,the value of $g(2)$ lies within the interval $[0, 2)$. Thus,$0 \le g(2) < 2$ is the correct inequality.

(A) We are given the integrals:
$\int_0^1 f(x) \, dx = 1$,
$\int_0^1 x f(x) \, dx = a$,
$\int_0^1 x^2 f(x) \, dx = a^2$.
We need to evaluate the integral $I = \int_0^1 (x - a)^2 f(x) \, dx$.
Expanding the term $(x - a)^2 = x^2 - 2ax + a^2$,we get:
$I = \int_0^1 (x^2 - 2ax + a^2) f(x) \, dx$
$I = \int_0^1 x^2 f(x) \, dx - 2a \int_0^1 x f(x) \, dx + a^2 \int_0^1 f(x) \, dx$
Substituting the given values:
$I = a^2 - 2a(a) + a^2(1)$
$I = a^2 - 2a^2 + a^2$
$I = 0$.

(C) Given $f(x) = A\sin \left( \frac{\pi x}{2} \right) + B$.
First,we use the integral condition: $\int_0^1 \left( A\sin \left( \frac{\pi x}{2} \right) + B \right) dx = \frac{2A}{\pi}$.
Evaluating the integral: $\left[ -\frac{2A}{\pi} \cos \left( \frac{\pi x}{2} \right) + Bx \right]_0^1 = \frac{2A}{\pi}$.
Substituting the limits: $\left( -\frac{2A}{\pi} \cos \left( \frac{\pi}{2} \right) + B(1) \right) - \left( -\frac{2A}{\pi} \cos(0) + B(0) \right) = \frac{2A}{\pi}$.
Since $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$,we get: $B - (-\frac{2A}{\pi}) = \frac{2A}{\pi} \implies B + \frac{2A}{\pi} = \frac{2A}{\pi} \implies B = 0$.
Now,find $f'(x)$: $f'(x) = A \cdot \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right)$.
Given $f'\left( \frac{1}{2} \right) = \sqrt{2}$,we have: $\frac{A\pi}{2} \cos \left( \frac{\pi}{4} \right) = \sqrt{2}$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we get: $\frac{A\pi}{2} \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \implies \frac{A\pi}{2} = 2 \implies A = \frac{4}{\pi}$.
Thus,$A = \frac{4}{\pi}$ and $B = 0$.

(C) Let the given integral be $J = \int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx}$.
Substitute $\lambda x = t$,which implies $x = \frac{t}{\lambda}$ and $dx = \frac{dt}{\lambda}$.
As $x \to 0$,$t \to 0$,and as $x \to \infty$,$t \to \infty$.
Substituting these into the integral:
$J = \int_0^\infty {{e^{ - t}}{(\frac{t}{\lambda})^{n - 1}} \frac{dt}{\lambda}}$
$J = \int_0^\infty {{e^{ - t}} \frac{{{t^{n - 1}}}}{{{\lambda ^{n - 1}}}} \frac{dt}{\lambda}}$
$J = \frac{1}{{{\lambda ^n}}} \int_0^\infty {{e^{ - t}}{t^{n - 1}}dt}$
Since ${I_n} = \int_0^\infty {{e^{ - x}}{x^{n - 1}}dx}$,the integral $\int_0^\infty {{e^{ - t}}{t^{n - 1}}dt}$ is also equal to ${I_n}$.
Therefore,$J = \frac{{{I_n}}}{{{\lambda ^n}}}$.

(B) Given $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$.
By the Fundamental Theorem of Calculus,$f'(x) = 6\cos x - 2\sin x$.
In the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,$\cos x > 0$ and $\sin x < 0$,so $6\cos x - 2\sin x > 0$.
Since $f'(x) > 0$,$f(x)$ is an increasing function on the given interval.
Therefore,the greatest value occurs at the upper limit $x = \frac{7\pi}{4}$.
$f\left( \frac{7\pi}{4} \right) = \int_{5\pi/3}^{7\pi/4} (6\cos t - 2\sin t) \, dt = [6\sin t + 2\cos t]_{5\pi/3}^{7\pi/4}$.
$= \left( 6\sin\frac{7\pi}{4} + 2\cos\frac{7\pi}{4} \right) - \left( 6\sin\frac{5\pi}{3} + 2\cos\frac{5\pi}{3} \right)$.
$= \left( 6\left(-\frac{1}{\sqrt{2}}\right) + 2\left(\frac{1}{\sqrt{2}}\right) \right) - \left( 6\left(-\frac{\sqrt{3}}{2}\right) + 2\left(\frac{1}{2}\right) \right)$.
$= \left( -\frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}} \right) - (-3\sqrt{3} + 1)$.
$= -\frac{4}{\sqrt{2}} + 3\sqrt{3} - 1 = 3\sqrt{3} - 2\sqrt{2} - 1$.

(D) For $0 < x < 1$,we have $x^2 > x^3$. Since the base $2 > 1$,the function $f(x) = 2^x$ is strictly increasing,so $2^{x^2} > 2^{x^3}$ for $0 < x < 1$.
Integrating both sides from $0$ to $1$,we get $\int_0^1 2^{x^2} dx > \int_0^1 2^{x^3} dx$,which implies ${I_1} > {I_2}$.
For $1 < x < 2$,we have $x^3 > x^2$. Similarly,$2^{x^3} > 2^{x^2}$ for $1 < x < 2$.
Integrating both sides from $1$ to $2$,we get $\int_1^2 2^{x^3} dx > \int_1^2 2^{x^2} dx$,which implies ${I_4} > {I_3}$.
Thus,the correct statement is ${I_1} > {I_2}$.

(B) Given the equation: $2f(x) - 3f\left( \frac{1}{x} \right) = x$ ... $(i)$
Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get:
$2f\left( \frac{1}{x} \right) - 3f(x) = \frac{1}{x}$ ... $(ii)$
To eliminate $f\left( \frac{1}{x} \right)$,multiply $(i)$ by $2$ and $(ii)$ by $3$:
$4f(x) - 6f\left( \frac{1}{x} \right) = 2x$
$6f\left( \frac{1}{x} \right) - 9f(x) = \frac{3}{x}$
Adding these two equations:
$-5f(x) = 2x + \frac{3}{x}$
$f(x) = -\frac{1}{5} \left( 2x + \frac{3}{x} \right)$
Now,integrate $f(x)$ from $1$ to $2$:
$\int_1^2 f(x) \, dx = -\frac{1}{5} \int_1^2 \left( 2x + \frac{3}{x} \right) \, dx$
$= -\frac{1}{5} \left[ x^2 + 3\ln|x| \right]_1^2$
$= -\frac{1}{5} \left( (2^2 + 3\ln 2) - (1^2 + 3\ln 1) \right)$
$= -\frac{1}{5} (4 + 3\ln 2 - 1 - 0)$
$= -\frac{1}{5} (3 + 3\ln 2) = -\frac{3}{5}(1 + \ln 2)$.

(D) Given $\int_a^b {x^3} dx = 0$.
Evaluating the integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \implies b^4 - a^4 = 0 \implies (b^2 - a^2)(b^2 + a^2) = 0$.
Since $a$ and $b$ are real,$b^2 + a^2 = 0$ implies $a=0, b=0$,which contradicts the second integral. Thus,$b^2 - a^2 = 0$,which means $b = a$ or $b = -a$.
If $b = a$,the integral $\int_a^b {x^2} dx = 0$,but we are given $\int_a^b {x^2} dx = \frac{2}{3}$. Thus,$b = -a$.
Given $\int_a^b {x^2} dx = \frac{2}{3}$,we have $\left[ \frac{x^3}{3} \right]_a^b = \frac{2}{3} \implies b^3 - a^3 = 2$.
Substituting $b = -a$: $(-a)^3 - a^3 = 2 \implies -2a^3 = 2 \implies a^3 = -1 \implies a = -1$.
Since $b = -a$,$b = -(-1) = 1$.
Therefore,$(a, b) = (-1, 1)$.

(C) Let $f(x) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$.
By the Fundamental Theorem of Calculus,the derivative is $f'(x) = x{e^{ - {x^2}}}$.
Setting $f'(x) = 0$,we get $x{e^{ - {x^2}}} = 0$. Since ${e^{ - {x^2}}} \neq 0$ for all real $x$,we have $x = 0$.
Now,find the second derivative: $f''(x) = {e^{ - {x^2}}} - 2{x^2}{e^{ - {x^2}}} = {e^{ - {x^2}}}(1 - 2{x^2})$.
Evaluating at $x = 0$,$f''(0) = {e^0}(1 - 0) = 1 > 0$.
Since the second derivative is positive at $x = 0$,the function has a local minimum at $x = 0$.
The minimum value is $f(0) = \int\limits_0^0 {t{e^{ - {t^2}}}} dt = 0$.

(B) Let $n = [a]$,where $n$ is an integer such that $n \leq a < n+1$.
The integral can be split as:
$\int_{1}^{a} [x] f'(x) dx = \int_{1}^{2} 1 f'(x) dx + \int_{2}^{3} 2 f'(x) dx + \dots + \int_{n-1}^{n} (n-1) f'(x) dx + \int_{n}^{a} n f'(x) dx$
Evaluating each integral:
$= 1[f(2) - f(1)] + 2[f(3) - f(2)] + \dots + (n-1)[f(n) - f(n-1)] + n[f(a) - f(n)]$
Rearranging the terms:
$= -f(1) - f(2) - f(3) - \dots - f(n) + n f(a)$
Since $n = [a]$,we get:
$= [a] f(a) - \{f(1) + f(2) + \dots + f([a])\}$

(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.
Using the standard integral formula $\int \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1} t$,we evaluate the definite integral:
$\left[\sec^{-1} t\right]_{\sqrt{2}}^{x} = \frac{\pi}{2}$.
Substituting the limits,we get:
$\sec^{-1} x - \sec^{-1} \sqrt{2} = \frac{\pi}{2}$.
Since $\sec^{-1} \sqrt{2} = \frac{\pi}{4}$,the equation becomes:
$\sec^{-1} x - \frac{\pi}{4} = \frac{\pi}{2}$.
Solving for $\sec^{-1} x$:
$\sec^{-1} x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Therefore,$x = \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}$.
Since $-\sqrt{2}$ is not among the given options,the correct choice is $(D)$.

(B) For $x \in (0, 1]$,we know that $\sin x < x$.
Therefore,$I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx < \int_{0}^{1} \frac{x}{\sqrt{x}} \, dx = \int_{0}^{1} x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3}$.
Thus,$I < \frac{2}{3}$.
For $x \in (0, 1]$,we know that $\cos x < 1$.
Therefore,$J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx < \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx = \int_{0}^{1} x^{-1/2} \, dx = \left[ 2x^{1/2} \right]_{0}^{1} = 2$.
Thus,$J < 2$.
Hence,the correct option is $B$.

(C) We need to evaluate the integral $I = \int_0^{1.5} x[x^2] dx$.
Since $[x^2]$ changes its value at $x^2 = 1$ and $x^2 = 2$,we split the integral at $x = 1$ and $x = \sqrt{2}$.
$I = \int_0^1 x[x^2] dx + \int_1^{\sqrt{2}} x[x^2] dx + \int_{\sqrt{2}}^{1.5} x[x^2] dx$.
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$1 \le x^2 < 2$,so $[x^2] = 1$.
For $\sqrt{2} \le x < 1.5$,$2 \le x^2 < 2.25$,so $[x^2] = 2$.
Thus,$I = \int_0^1 x(0) dx + \int_1^{\sqrt{2}} x(1) dx + \int_{\sqrt{2}}^{1.5} x(2) dx$.
$I = 0 + \left[ \frac{x^2}{2} \right]_1^{\sqrt{2}} + 2 \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{1.5}$.
$I = \left( \frac{2-1}{2} \right) + (1.5^2 - (\sqrt{2})^2) = \frac{1}{2} + (2.25 - 2) = 0.5 + 0.25 = 0.75 = \frac{3}{4}$.

(D) Given $g(x) = \int_{0}^{x} \cos 4t \, dt$.
Evaluating the integral,we get $g(x) = \left[ \frac{\sin 4t}{4} \right]_{0}^{x} = \frac{\sin 4x}{4}$.
Now,calculate $g(x + \pi) = \frac{\sin 4(x + \pi)}{4} = \frac{\sin(4x + 4\pi)}{4}$.
Since $\sin(4x + 4\pi) = \sin 4x$,we have $g(x + \pi) = \frac{\sin 4x}{4} = g(x)$.
Also,$g(\pi) = \frac{\sin(4\pi)}{4} = 0$.
Since $g(\pi) = 0$,it follows that $g(x) + g(\pi) = g(x) + 0 = g(x)$ and $g(x) - g(\pi) = g(x) - 0 = g(x)$.
Thus,both $g(x) + g(\pi)$ and $g(x) - g(\pi)$ are equal to $g(x)$.

(B) The given integral is $I = \int_0^\pi \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \; dx$.
This can be written as $I = \int_0^\pi \sqrt{(1 - 2\sin\frac{x}{2})^2} \; dx = \int_0^\pi |1 - 2\sin\frac{x}{2}| \; dx$.
We know $1 - 2\sin\frac{x}{2} = 0$ when $\sin\frac{x}{2} = \frac{1}{2}$,which means $\frac{x}{2} = \frac{\pi}{6}$,so $x = \frac{\pi}{3}$.
For $0 \le x < \frac{\pi}{3}$,$1 - 2\sin\frac{x}{2} > 0$. For $\frac{\pi}{3} < x \le \pi$,$1 - 2\sin\frac{x}{2} < 0$.
Thus,$I = \int_0^{\pi/3} (1 - 2\sin\frac{x}{2}) \; dx + \int_{\pi/3}^\pi -(1 - 2\sin\frac{x}{2}) \; dx$.
$I = [x + 4\cos\frac{x}{2}]_0^{\pi/3} - [x + 4\cos\frac{x}{2}]_{\pi/3}^\pi$.
$I = ((\frac{\pi}{3} + 4\cos\frac{\pi}{6}) - (0 + 4\cos 0)) - ((\pi + 4\cos\frac{\pi}{2}) - (\frac{\pi}{3} + 4\cos\frac{\pi}{6}))$.
$I = (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2}) - 4) - ((\pi + 0) - (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2})))$.
$I = (\frac{\pi}{3} + 2\sqrt{3} - 4) - (\pi - \frac{\pi}{3} - 2\sqrt{3}) = \frac{\pi}{3} + 2\sqrt{3} - 4 - \frac{2\pi}{3} + 2\sqrt{3} = 4\sqrt{3} - 4 - \frac{\pi}{3}$.

(A) We are given $g(2) = \int_0^2 f(t) dt = \int_0^1 f(t) dt + \int_1^2 f(t) dt$.
For $t \in [0, 1]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating this,we get $\int_0^1 0 dt \le \int_0^1 f(t) dt \le \int_0^1 \frac{1}{2} dt$,which simplifies to $0 \le \int_0^1 f(t) dt \le \frac{1}{2}$.
For $t \in [1, 2]$,we have $\frac{1}{2} \le f(t) \le 1$. Integrating this,we get $\int_1^2 \frac{1}{2} dt \le \int_1^2 f(t) dt \le \int_1^2 1 dt$,which simplifies to $\frac{1}{2} \le \int_1^2 f(t) dt \le 1$.
Adding these two inequalities,we get $0 + \frac{1}{2} \le \int_0^1 f(t) dt + \int_1^2 f(t) dt \le \frac{1}{2} + 1$.
Therefore,$\frac{1}{2} \le g(2) \le \frac{3}{2}$.