Properties of definite integration Questions in English

(C) Let $I = \int_0^\pi e^{\cos^2 x} \cos^5(3x) \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^\pi e^{\cos^2(\pi-x)} \cos^5(3(\pi-x)) \,dx$
Since $\cos(\pi-x) = -\cos x$,we have $\cos^2(\pi-x) = \cos^2 x$.
Also,$\cos(3(\pi-x)) = \cos(3\pi - 3x) = \cos(3\pi) \cos(3x) + \sin(3\pi) \sin(3x) = -\cos(3x)$.
Therefore,$\cos^5(3(\pi-x)) = (-\cos(3x))^5 = -\cos^5(3x)$.
Substituting these into the integral:
$I = \int_0^\pi e^{\cos^2 x} (-\cos^5(3x)) \,dx = -\int_0^\pi e^{\cos^2 x} \cos^5(3x) \,dx = -I$.
Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

(B) Let $I = \int_0^{\pi /2} \frac{1}{1 + \sqrt{\tan x}} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we have:
$I = \int_0^{\pi /2} \frac{1}{1 + \sqrt{\tan(\pi/2 - x)}} \,dx = \int_0^{\pi /2} \frac{1}{1 + \sqrt{\cot x}} \,dx$.
$I = \int_0^{\pi /2} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} \,dx = \int_0^{\pi /2} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + 1} \,dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi /2} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \right) \,dx$.
$2I = \int_0^{\pi /2} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} \,dx = \int_0^{\pi /2} 1 \,dx$.
$2I = [x]_0^{\pi /2} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(C) Let $I = \int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}} \,dx$.
We can split the integral as:
$I = \int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx - \int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$.
Let $f(x) = \frac{{\sin x}}{{3 - |x|}}$ and $g(x) = \frac{{{x^2}}}{{3 - |x|}}$.
For $f(x)$,$f(-x) = \frac{{\sin(-x)}}{{3 - |-x|}} = \frac{{-\sin x}}{{3 - |x|}} = -f(x)$,so $f(x)$ is an odd function.
Thus,$\int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx = 0$.
For $g(x)$,$g(-x) = \frac{{(-x)^2}}{{3 - |-x|}} = \frac{{{x^2}}}{{3 - |x|}} = g(x)$,so $g(x)$ is an even function.
Thus,$\int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$.
Substituting these back into $I$:
$I = 0 - 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$.

(D) Let $f(x) = \sin^{11} x$.
To check if the function is even or odd,we evaluate $f(-x)$:
$f(-x) = \sin^{11}(-x) = (\sin(-x))^{11} = (-\sin x)^{11} = -\sin^{11} x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x) = \sin^{11} x$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} \sin^{11} x \, dx = 0$.

(D) Let the given integral be $I = \int_{-2}^{2} (px^2 + qx + s) \, dx$.
We can split this into three integrals:
$I = p \int_{-2}^{2} x^2 \, dx + q \int_{-2}^{2} x \, dx + s \int_{-2}^{2} 1 \, dx$.
Since $f(x) = x$ is an odd function,$\int_{-2}^{2} x \, dx = 0$.
Since $f(x) = x^2$ and $f(x) = 1$ are even functions,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2p \int_{0}^{2} x^2 \, dx + 0 + 2s \int_{0}^{2} 1 \, dx$.
$I = 2p \left[ \frac{x^3}{3} \right]_{0}^{2} + 2s [x]_{0}^{2}$.
$I = 2p \left( \frac{8}{3} \right) + 2s(2) = \frac{16p}{3} + 4s$.
Therefore,to find the numerical value of $I$,it is necessary to know the values of $p$ and $s$.

(C) Let $f(x) = \log \left( \frac{1+x}{1-x} \right)$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left( \frac{1+(-x)}{1-(-x)} \right) = \log \left( \frac{1-x}{1+x} \right) = \log \left( \left( \frac{1+x}{1-x} \right)^{-1} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = 0$.

(A) Let $f(t) = \log \left( \frac{1 - t}{1 + t} \right)$.
First,we check if $f(t)$ is an odd function by evaluating $f(-t)$:
$f(-t) = \log \left( \frac{1 - (-t)}{1 + (-t)} \right) = \log \left( \frac{1 + t}{1 - t} \right) = \log \left( \left( \frac{1 - t}{1 + t} \right)^{-1} \right) = -\log \left( \frac{1 - t}{1 + t} \right) = -f(t)$.
Since $f(-t) = -f(t)$,the function $f(t)$ is an odd function.
We know that if $f(t)$ is an odd function,then the integral $F(x) = \int_0^x f(t) \,dt$ is an even function.
To verify,$F(-x) = \int_0^{-x} f(t) \,dt$. Let $t = -u$,then $dt = -du$. When $t=0, u=0$; when $t=-x, u=x$.
$F(-x) = \int_0^x f(-u) (-du) = \int_0^x (-f(u)) (-du) = \int_0^x f(u) \,du = F(x)$.
Since $F(-x) = F(x)$,$F(x)$ is an even function.

(A) Let $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}dx}$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - \pi /2 + \pi /2 - x)}}{{1 + {e^{ - \pi /2 + \pi /2 - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - x)}}{{1 + {e^{ - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^{ - x}}}}dx$.
Since $\cos(-x) = \cos x$,we have $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + 1/e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{e^x + 1}}}dx$.
Adding the two expressions for $I$:
$2I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^x}}}dx + \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x(1 + e^x)}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\cos x} dx$.
$2I = [\sin x]_{ - \pi /2}^{\pi /2} = \sin(\pi /2) - \sin( - \pi /2) = 1 - (-1) = 2$.
Therefore,$I = 1$.

(C) Let $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin(\pi/2 - x)}}{{4 + 3\cos(\pi/2 - x)}}} \right)} \,dx$.
Since $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$,we have:
$I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \,dx$.
Using the property $\log(a/b) = -\log(b/a)$,we get:
$I = \int_0^{\pi /2} -\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \,dx = -I$.
Therefore,$2I = 0$,which implies $I = 0$.

(D) Let $I = \int_0^{2\pi } {{\cos }^{99}x\,dx}$.
Using the property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
Since $\cos^{99}(2\pi - x) = \cos^{99}x$,we have $I = 2\int_0^{\pi} \cos^{99}x\,dx$.
Using the property $\int_0^a f(x) dx = 0$ if $f(a-x) = -f(x)$:
Since $\cos^{99}(\pi - x) = -\cos^{99}x$,the integral $\int_0^{\pi} \cos^{99}x\,dx = 0$.
Therefore,$I = 2 \times 0 = 0$.

(B) Let $I = \int_{0}^{\pi} \cos^3 x \, dx$.
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we have:
$I = \int_{0}^{\pi} \cos^3(\pi - x) \, dx$.
Since $\cos(\pi - x) = -\cos x$,then $\cos^3(\pi - x) = -\cos^3 x$.
So,$I = \int_{0}^{\pi} -\cos^3 x \, dx = -I$.
$2I = 0 \implies I = 0$.
Alternatively,using the property $\int_{0}^{2a} f(x) \, dx = 0$ if $f(2a-x) = -f(x)$,here $f(x) = \cos^3 x$ and $2a = \pi$,so $f(\pi - x) = -f(x)$,thus the integral is $0$.

(A) Let $f(x) = \frac{{{x^2}\sin 2x}}{{{x^2} + 1}}$.
Check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \frac{{{{( - x)}^2}\sin (2( - x))}}{{{{( - x)}^2} + 1}} = \frac{{{x^2}\sin ( - 2x)}}{{{x^2} + 1}} = \frac{{ - {x^2}\sin 2x}}{{{x^2} + 1}} = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{ - a}^a {f(x)\,dx = 0}$.
Therefore,$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = 0}$.

(A) Let $I = \int_{0}^{\pi} \log(\sin^2 x) \, dx$.
Using the property $\log(a^b) = b \log a$,we have $I = \int_{0}^{\pi} 2 \log(\sin x) \, dx$.
Since $\sin x$ is symmetric about $x = \frac{\pi}{2}$ in the interval $[0, \pi]$,we can write $I = 2 \times 2 \int_{0}^{\pi/2} \log(\sin x) \, dx = 4 \int_{0}^{\pi/2} \log(\sin x) \, dx$.
Using the standard integral result $\int_{0}^{\pi/2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2$,we get:
$I = 4 \times \left(-\frac{\pi}{2} \log 2\right) = -2\pi \log 2$.
Since $-\log 2 = \log(2^{-1}) = \log(\frac{1}{2})$,we have $I = 2\pi \log(\frac{1}{2})$.

(C) Let $I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $.
Since $\cos(-x) = \cos x$,the function $g(x) = f(\cos x)$ is an even function because $g(-x) = f(\cos(-x)) = f(\cos x) = g(x)$.
For an even function,the property $\int_{-a}^{a} g(x) \, dx = 2 \int_{0}^{a} g(x) \, dx$ holds.
Thus,$I = 2 \int_{0}^{\frac{\pi }{2}} f(\cos x) \, dx$.
Using the property $\int_{0}^{a} h(x) \, dx = \int_{0}^{a} h(a-x) \, dx$,we substitute $x = \frac{\pi}{2} - t$:
$I = 2 \int_{0}^{\frac{\pi }{2}} f(\cos(\frac{\pi}{2} - t)) \, dt = 2 \int_{0}^{\frac{\pi }{2}} f(\sin t) \, dt$.
Therefore,the correct option is $C$.

(B) Let $I = \int_0^\pi \sin^2 x \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a - x) = f(x)$:
Here,$f(x) = \sin^2 x$ and $2a = \pi$,so $a = \frac{\pi}{2}$.
Since $f(\pi - x) = \sin^2(\pi - x) = \sin^2 x = f(x)$,the property holds.
Thus,$I = 2 \int_0^{\pi/2} \sin^2 x \, dx$.
Using the Wallis formula or $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$I = 2 \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx = \int_0^{\pi/2} (1 - \cos 2x) \, dx$.
$I = [x - \frac{\sin 2x}{2}]_0^{\pi/2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$.

(C) Let $I = \int_0^{\pi /2} \frac{\sin x}{\sin x + \cos x} \, dx$ --- $(1)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx$
$I = \int_0^{\pi /2} \frac{\cos x}{\cos x + \sin x} \, dx$ --- $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^{\pi /2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx$
$2I = \int_0^{\pi /2} 1 \, dx$
$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(A) Let $I = \int_{-1}^{1} x \tan^{-1} x \, dx$.
Since $f(x) = x \tan^{-1} x$ is an even function (because $f(-x) = (-x) \tan^{-1}(-x) = (-x)(-\tan^{-1} x) = x \tan^{-1} x = f(x)$),we can write:
$I = 2 \int_{0}^{1} x \tan^{-1} x \, dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$.
$I = 2 \left[ \frac{x^2}{2} \tan^{-1} x \Big|_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx \right]$
$I = \left[ x^2 \tan^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{1+x^2} \, dx$
$I = (1^2 \tan^{-1} 1 - 0) - \int_0^1 \frac{x^2+1-1}{1+x^2} \, dx$
$I = \frac{\pi}{4} - \int_0^1 \left( 1 - \frac{1}{1+x^2} \right) \, dx$
$I = \frac{\pi}{4} - [x - \tan^{-1} x]_0^1$
$I = \frac{\pi}{4} - (1 - \tan^{-1} 1) = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1$.

(B) Let $I = \int_{-a}^{a} \sin x \, f(\cos x) \, dx$.
Define $g(x) = \sin x \, f(\cos x)$.
Then $g(-x) = \sin(-x) \, f(\cos(-x)) = -\sin x \, f(\cos x) = -g(x)$.
Since $g(-x) = -g(x)$,the function $g(x)$ is an odd function.
According to the property of definite integrals,if $g(x)$ is an odd function,then $\int_{-a}^{a} g(x) \, dx = 0$.
Therefore,$I = 0$.

(A) Let $I = \int_{0}^{\pi /2} \frac{2^{\sin x}}{2^{\sin x} + 2^{\cos x}} dx$ ....$(i)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi /2} \frac{2^{\sin(\pi/2 - x)}}{2^{\sin(\pi/2 - x)} + 2^{\cos(\pi/2 - x)}} dx$
Since $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$,we have:
$I = \int_{0}^{\pi /2} \frac{2^{\cos x}}{2^{\cos x} + 2^{\sin x}} dx$ ....$(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi /2} \left( \frac{2^{\sin x} + 2^{\cos x}}{2^{\sin x} + 2^{\cos x}} \right) dx$
$2I = \int_{0}^{\pi /2} 1 dx$
$2I = [x]_{0}^{\pi /2} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$.

(C) Let $I = \int_{ - \pi /2}^{\pi /2} {f(x)dx} $,where $f(x) = \frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}$.
Check if $f(x)$ is an odd or even function:
$f(-x) = \frac{{\sin(-x)}}{{1 + {{\cos }^2}(-x)}}{e^{ - {{\cos }^2}(-x)}}$
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = \frac{{-\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}} = -f(x)$.
Since $f(x)$ is an odd function,by the property of definite integrals $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is odd,we get:
$I = 0$.

(B) Let $I = \int_{0.5}^{1.5} xf(x) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have:
$I = \int_{0.5}^{1.5} (0.5 + 1.5 - x) f(0.5 + 1.5 - x) dx$
$I = \int_{0.5}^{1.5} (2 - x) f(2 - x) dx$.
Since $f(2 - x) = f(x)$,we substitute this into the integral:
$I = \int_{0.5}^{1.5} (2 - x) f(x) dx$
$I = 2 \int_{0.5}^{1.5} f(x) dx - \int_{0.5}^{1.5} xf(x) dx$
$I = 2 \int_{0.5}^{1.5} f(x) dx - I$
$2I = 2 \int_{0.5}^{1.5} f(x) dx$
$I = \int_{0.5}^{1.5} f(x) dx$.

(A) Let $I = \int_{0}^{\pi /2} \frac{e^{x^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi /2} \frac{e^{(\pi /2 - x)^2}}{e^{(\pi /2 - x)^2} + e^{x^2}} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi /2} \frac{e^{x^2} + e^{(\pi /2 - x)^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$.
$2I = \int_{0}^{\pi /2} 1 dx = [x]_{0}^{\pi /2} = \pi /2$.
Therefore,$I = \pi /4$.

(D) Let $I = \int_{-2}^{2} |x| \, dx$.
Since $|x|$ is an even function,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
$I = 2 \int_{0}^{2} x \, dx$.
$I = 2 \left[ \frac{x^2}{2} \right]_{0}^{2}$.
$I = 2 \left( \frac{2^2}{2} - \frac{0^2}{2} \right)$.
$I = 2 \left( \frac{4}{2} \right) = 2(2) = 4$.

(D) Given that $f(-x) = -f(x)$,the function $f$ is an odd function.
For an odd function,the property of definite integrals states that $\int_{-a}^{a} f(x) dx = 0$.
We can split the integral as: $\int_{-1}^{1} f(x) dx = \int_{-1}^{0} f(x) dx + \int_{0}^{1} f(x) dx = 0$.
Substituting the given value $\int_{0}^{1} f(x) dx = 5$,we get: $\int_{-1}^{0} f(x) dx + 5 = 0$.
Therefore,$\int_{-1}^{0} f(x) dx = -5$.
Since the variable of integration is a dummy variable,$\int_{-1}^{0} f(t) dt = \int_{-1}^{0} f(x) dx = -5$.

(C) Given ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$.
Using the property $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }$,we have:
${I_1} = \int_a^{\pi - a} {(\pi - x)f(\sin(\pi - x))dx}$.
Since $\sin(\pi - x) = \sin x$,this becomes:
${I_1} = \int_a^{\pi - a} {(\pi - x)f(\sin x)dx}$.
${I_1} = \pi \int_a^{\pi - a} {f(\sin x)dx} - \int_a^{\pi - a} {xf(\sin x)dx}$.
${I_1} = \pi {I_2} - {I_1}$.
$2{I_1} = \pi {I_2}$.
Therefore,${I_2} = \frac{2}{\pi }{I_1}$.

(A) Let $I = \int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$.
Define $f(x) = \cos x \ln \left( \frac{1 + x}{1 - x} \right)$.
Check if $f(x)$ is an odd function by evaluating $f(-x)$:
$f(-x) = \cos(-x) \ln \left( \frac{1 + (-x)}{1 - (-x)} \right) = \cos x \ln \left( \frac{1 - x}{1 + x} \right)$.
Since $\ln \left( \frac{1 - x}{1 + x} \right) = \ln \left( \frac{1 + x}{1 - x} \right)^{-1} = -\ln \left( \frac{1 + x}{1 - x} \right)$,we have $f(-x) = -\cos x \ln \left( \frac{1 + x}{1 - x} \right) = -f(x)$.
Thus,$f(x)$ is an odd function.
By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.
Therefore,$I = 0$.

(B) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$.
Since $\log_e x < 0$ for $x \in [e^{-1}, 1)$ and $\log_e x \ge 0$ for $x \in [1, e^2]$,we split the integral:
$I = \int_{e^{-1}}^{1} -\frac{\log_e x}{x} dx + \int_{1}^{e^2} \frac{\log_e x}{x} dx$.
Let $z = \log_e x$,then $dz = \frac{1}{x} dx$.
When $x = e^{-1}, z = -1$. When $x = 1, z = 0$. When $x = e^2, z = 2$.
$I = \int_{-1}^{0} -z dz + \int_{0}^{2} z dz$.
$I = \left[ -\frac{z^2}{2} \right]_{-1}^{0} + \left[ \frac{z^2}{2} \right]_{0}^{2}$.
$I = (0 - (-1/2)) + (4/2 - 0) = \frac{1}{2} + 2 = \frac{5}{2}$.

(C) We are given the function $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \\ 2, & \text{otherwise} \end{cases}$.
We need to evaluate the integral $I = \int_{-2}^{3} f(x) dx$.
We can split the integral at $x = 2$ as follows:
$I = \int_{-2}^{2} f(x) dx + \int_{2}^{3} f(x) dx$.
For the interval $[-2, 2]$,$f(x) = e^{\cos x}\sin x$. Let $g(x) = e^{\cos x}\sin x$. Then $g(-x) = e^{\cos(-x)}\sin(-x) = e^{\cos x}(-\sin x) = -g(x)$. Thus,$g(x)$ is an odd function.
By the property of definite integrals,if $g(x)$ is an odd function,then $\int_{-a}^{a} g(x) dx = 0$. Therefore,$\int_{-2}^{2} e^{\cos x}\sin x dx = 0$.
For the interval $(2, 3]$,$f(x) = 2$. Thus,$\int_{2}^{3} 2 dx = [2x]_{2}^{3} = 2(3 - 2) = 2$.
Adding these results,$I = 0 + 2 = 2$.

(A) Let $\phi(x) = [f(x) + f(-x)][g(x) - g(-x)]$.
Now,calculate $\phi(-x)$:
$\phi(-x) = [f(-x) + f(x)][g(-x) - g(x)]$
$\phi(-x) = [f(x) + f(-x)][-(g(x) - g(-x))]$
$\phi(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -\phi(x)$.
Since $\phi(-x) = -\phi(x)$,the function $\phi(x)$ is an odd function.
For any odd function $\phi(x)$,the integral over the symmetric interval $[-\pi, \pi]$ is zero:
$\int_{-\pi}^{\pi} \phi(x) \, dx = 0$.
Therefore,$\int_{-\pi}^{\pi} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

(C) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.
We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ....$(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \pi/6$ and $b = \pi/3$,we have $a+b = \pi/2$.
Thus,$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ....(ii)
Adding $(i)$ and (ii):
$2I = \int_{\pi /6}^{\pi /3} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx = \int_{\pi /6}^{\pi /3} 1 dx$
$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
Therefore,$I = \frac{1}{2} \times \frac{\pi}{6} = \frac{\pi}{12}$.

(A) Let $I = \int_{0}^{\pi /2} \frac{\sin^{2/3} x}{\sin^{2/3} x + \cos^{2/3} x} dx$ $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi /2} \frac{\sin^{2/3}(\pi /2 - x)}{\sin^{2/3}(\pi /2 - x) + \cos^{2/3}(\pi /2 - x)} dx$
$I = \int_{0}^{\pi /2} \frac{\cos^{2/3} x}{\cos^{2/3} x + \sin^{2/3} x} dx$ $(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\pi /2} \frac{\sin^{2/3} x + \cos^{2/3} x}{\sin^{2/3} x + \cos^{2/3} x} dx$
$2I = \int_{0}^{\pi /2} 1 dx$
$2I = [x]_{0}^{\pi /2} = \pi /2$
$I = \pi /4$
General property: $\int_{0}^{\pi /2} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} dx = \frac{\pi}{4}$.

(A) Let $f(x) = \log(x + \sqrt{1 + x^2})$.
Now,$f(-x) = \log(-x + \sqrt{1 + (-x)^2}) = \log(\sqrt{1 + x^2} - x)$.
Multiplying and dividing by $(\sqrt{1 + x^2} + x)$,we get:
$f(-x) = \log\left(\frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x}\right)$
$= \log\left(\frac{(1 + x^2) - x^2}{\sqrt{1 + x^2} + x}\right)$
$= \log\left(\frac{1}{\sqrt{1 + x^2} + x}\right)$
$= \log(1) - \log(\sqrt{1 + x^2} + x) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
By the property of definite integrals,$\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.
Therefore,$\int_{-1}^{1} \log(x + \sqrt{1 + x^2}) \, dx = 0$.

(C) By the property of definite integrals,we can split the interval $[0, 2a]$ into $[0, a]$ and $[a, 2a]$.
Thus,$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx$.
In the second integral,let $x = 2a - t$,then $dx = -dt$.
When $x = a$,$t = a$,and when $x = 2a$,$t = 0$.
So,$\int_{a}^{2a} f(x) \, dx = \int_{a}^{0} f(2a - t) (-dt) = \int_{0}^{a} f(2a - t) \, dt = \int_{0}^{a} f(2a - x) \, dx$.
Therefore,$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx$.

(B) Let $I = \int_{0}^{\pi} e^{\sin^2 x} \cos^3 x \, dx$ ... $(i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} e^{\sin^2(\pi - x)} \cos^3(\pi - x) \, dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have:
$I = \int_{0}^{\pi} e^{\sin^2 x} (-\cos x)^3 \, dx$
$I = - \int_{0}^{\pi} e^{\sin^2 x} \cos^3 x \, dx$ ... $(ii)$
Adding $(i)$ and $(ii)$,we get:
$I + I = \int_{0}^{\pi} e^{\sin^2 x} \cos^3 x \, dx - \int_{0}^{\pi} e^{\sin^2 x} \cos^3 x \, dx$
$2I = 0$
$I = 0$

(C) Let $I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx$ .....$(i)$
Using the property $\int_{A}^{B} f(x) dx = \int_{A}^{B} f(A + B - x) dx$,where $A = \frac{1}{n}$ and $B = \frac{an - 1}{n}$,we have $A + B = \frac{1 + an - 1}{n} = a$.
So,$I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{a - x}}{\sqrt{a - (a - x)} + \sqrt{a - x}} dx = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} dx$ .....$(ii)$
Adding $(i)$ and $(ii)$,we get:
$2I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \left( \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} + \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} \right) dx$
$2I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} 1 dx$
$2I = [x]_{\frac{1}{n}}^{\frac{an - 1}{n}}$
$2I = \frac{an - 1}{n} - \frac{1}{n} = \frac{an - 2}{n}$
$I = \frac{an - 2}{2n}$

(C) Let $I = \int_{0}^{\pi /2} {\sin 2x \log \tan x \, dx}$.
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi /2} {\sin 2(\frac{\pi}{2} - x) \log \tan(\frac{\pi}{2} - x) \, dx}$
$I = \int_{0}^{\pi /2} {\sin(\pi - 2x) \log \cot x \, dx}$
Since $\sin(\pi - 2x) = \sin 2x$ and $\log \cot x = \log(\tan x)^{-1} = -\log \tan x$,we have:
$I = \int_{0}^{\pi /2} {\sin 2x (-\log \tan x) \, dx}$
$I = - \int_{0}^{\pi /2} {\sin 2x \log \tan x \, dx}$
$I = -I$
$2I = 0 \Rightarrow I = 0$.

(A) Let $I = \int_{-1/2}^{1/2} [x] dx + \int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx$.
Consider the function $f(x) = \log \left( \frac{1+x}{1-x} \right)$.
Since $f(-x) = \log \left( \frac{1-x}{1+x} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$,$f(x)$ is an odd function.
Therefore,$\int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx = 0$.
Now,$I = \int_{-1/2}^{1/2} [x] dx$.
In the interval $[-1/2, 0)$,$[x] = -1$.
In the interval $[0, 1/2]$,$[x] = 0$.
Thus,$I = \int_{-1/2}^{0} (-1) dx + \int_{0}^{1/2} (0) dx = -[x]_{-1/2}^{0} = -(0 - (-1/2)) = -1/2$.

(C) Let $I = \int_{-\pi/2}^{\pi/2} (3\sin x + \sin^3 x) \, dx$.
Consider the function $f(x) = 3\sin x + \sin^3 x$.
Check if the function is even or odd by substituting $-x$ for $x$:
$f(-x) = 3\sin(-x) + \sin^3(-x)$
Since $\sin(-x) = -\sin x$,we have:
$f(-x) = 3(-\sin x) + (-\sin x)^3 = -3\sin x - \sin^3 x = -(3\sin x + \sin^3 x) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-\pi/2}^{\pi/2} (3\sin x + \sin^3 x) \, dx = 0$.

(D) Let $I = \int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$.
Using the identity $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right)$,we can write:
$\frac{1}{x^2 - x + 1} = \frac{x - (x - 1)}{1 + x(x - 1)}$.
Thus,$\tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) = \tan^{-1} x - \tan^{-1} (x - 1)$.
$I = \int_{0}^{1} \tan^{-1} x dx - \int_{0}^{1} \tan^{-1} (x - 1) dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$ for the second integral:
$\int_{0}^{1} \tan^{-1} (x - 1) dx = \int_{0}^{1} \tan^{-1} ((1 - x) - 1) dx = \int_{0}^{1} \tan^{-1} (-x) dx = -\int_{0}^{1} \tan^{-1} x dx$.
Substituting this back into $I$:
$I = \int_{0}^{1} \tan^{-1} x dx - (-\int_{0}^{1} \tan^{-1} x dx) = 2 \int_{0}^{1} \tan^{-1} x dx$.
Using integration by parts: $\int \tan^{-1} x dx = x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2)$.
$I = 2 [x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2)]_{0}^{1} = 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln 2) - (0 - 0) ] = 2 [ \frac{\pi}{4} - \frac{1}{2} \ln 2 ] = \frac{\pi}{2} - \ln 2$.

(C) Let $f(x) = p \ln \left( \frac{1+x}{1-x} \right) + q \ln \left( \frac{1-x}{1+x} \right)^{-2} + r$.
Note that $\ln \left( \frac{1-x}{1+x} \right)^{-2} = -2 \ln \left( \frac{1-x}{1+x} \right) = 2 \ln \left( \frac{1+x}{1-x} \right)$.
Thus,$f(x) = (p + 2q) \ln \left( \frac{1+x}{1-x} \right) + r$.
Since $\ln \left( \frac{1+x}{1-x} \right)$ is an odd function,the integral of this term over the symmetric interval $[-2, 2]$ is $0$.
Therefore,$\int_{-2}^{2} f(x) dx = \int_{-2}^{2} r dx = r [x]_{-2}^{2} = r(2 - (-2)) = 4r$.
Thus,the value of the integral depends only on $r$.

(C) Let $I = \int_0^\pi \frac{x \, dx}{1 + \sin x}$ ....$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:
$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin(\pi - x)}$
$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin x}$ ....$(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{x + \pi - x}{1 + \sin x} \, dx = \pi \int_0^\pi \frac{1}{1 + \sin x} \, dx$
Multiply numerator and denominator by $(1 - \sin x)$:
$2I = \pi \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} \, dx = \pi \int_0^\pi \frac{1 - \sin x}{\cos^2 x} \, dx$
$2I = \pi \int_0^\pi (\sec^2 x - \sec x \tan x) \, dx$
Integrating the terms:
$2I = \pi [\tan x - \sec x]_0^\pi$
$2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)]$
$2I = \pi [(0 - (-1)) - (0 - 1)]$
$2I = \pi [1 + 1] = 2\pi$
$I = \pi$

(B) Let $I = \int_0^\pi {xf(\sin x)dx}$.
Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we have:
$I = \int_0^\pi {(\pi - x)f(\sin(\pi - x))dx} = \int_0^\pi {(\pi - x)f(\sin x)dx}$.
Adding the two expressions for $I$:
$2I = \int_0^\pi {xf(\sin x)dx} + \int_0^\pi {(\pi - x)f(\sin x)dx} = \int_0^\pi {\pi f(\sin x)dx} = \pi \int_0^\pi {f(\sin x)dx}$.
Using the property $\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(2a-x) = f(x)$,we have $\int_0^\pi {f(\sin x)dx} = 2\int_0^{\pi /2} {f(\sin x)dx}$.
Thus,$2I = \pi \times 2 \int_0^{\pi /2} {f(\sin x)dx} = 2\pi \int_0^{\pi /2} {f(\sin x)dx}$.
Therefore,$I = \pi \int_0^{\pi /2} {f(\sin x)dx}$.
Comparing this with $A \int_0^{\pi /2} {f(\sin x)dx}$,we get $A = \pi$.

(C) Given $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$.
Using the property of definite integrals,$\int_0^{na} f(x) dx = n \int_0^a f(x) dx$ if $f(x+a) = f(x)$.
Here,$f(\cos^2(x+\pi)) = f(\cos^2 x)$ because $\cos^2(x+\pi) = (-\cos x)^2 = \cos^2 x$.
Thus,the function $f(\cos^2 x)$ is periodic with period $\pi$.
Therefore,$P = \int_0^{3\pi} f(\cos^2 x) dx = 3 \int_0^{\pi} f(\cos^2 x) dx = 3Q$.
This implies $P = 3Q$,or $P - 3Q = 0$.

(A) Let $I = \int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:
$I = \int_0^{2a} \frac{f(2a - x)}{f(2a - x) + f(2a - (2a - x))} \, dx$
$I = \int_0^{2a} \frac{f(2a - x)}{f(2a - x) + f(x)} \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{2a} \left( \frac{f(x)}{f(x) + f(2a - x)} + \frac{f(2a - x)}{f(2a - x) + f(x)} \right) \, dx$
$2I = \int_0^{2a} \frac{f(x) + f(2a - x)}{f(x) + f(2a - x)} \, dx$
$2I = \int_0^{2a} 1 \, dx$
$2I = [x]_0^{2a} = 2a$
$I = a$

(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$.
Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n \int_0^{\pi} \sin x \, dx = n [-\cos x]_0^{\pi} = n(1 - (-1)) = 2n$.
Now,consider the integral $\int_{n\pi}^{n\pi + v} |\sin x| \, dx$. Let $x = n\pi + t$,then $dx = dt$. When $x = n\pi$,$t = 0$. When $x = n\pi + v$,$t = v$.
Since $|sin(n\pi + t)| = |\sin(n\pi) \cos t + \cos(n\pi) \sin t| = |(-1)^n \sin t| = |\sin t|$.
So,$\int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_0^v |\sin t| \, dt$.
Assuming $0 \le v \le \pi$,we have $\int_0^v \sin t \, dt = [-\cos t]_0^v = 1 - \cos v$.
Therefore,$\int_0^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v$.

(A) Given ${u_n} = \int_0^{\pi /4} {{\tan ^n}x\,dx} $.
We can write ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{\tan ^n}x\,dx} + \int_0^{\pi /4} {{\tan ^{n - 2}}x\,dx} $.
$= \int_0^{\pi /4} {{\tan ^{n - 2}}x(\tan ^2x + 1)\,dx} $.
Since $\tan ^2x + 1 = \sec ^2x$,we have:
$= \int_0^{\pi /4} {{\tan ^{n - 2}}x\sec ^2x\,dx} $.
Let $t = \tan x$,then $dt = \sec ^2x\,dx$.
When $x = 0, t = 0$ and when $x = \pi /4, t = 1$.
$= \int_0^1 {{t^{n - 2}}\,dt} $.
$= \left[ \frac{t^{n - 1}}{n - 1} \right]_0^1 = \frac{1}{n - 1} - 0 = \frac{1}{n - 1}$.
Thus,${u_n} + {u_{n - 2}} = \frac{1}{n - 1}$.

(A) Let $I = \int_0^1 \log \sin \left( \frac{\pi }{2}x \right) dx$.
Substitute $\frac{\pi }{2}x = \theta$,then $dx = \frac{2}{\pi} d\theta$.
When $x = 0$,$\theta = 0$ and when $x = 1$,$\theta = \frac{\pi}{2}$.
Thus,$I = \frac{2}{\pi} \int_0^{\pi/2} \log \sin \theta \, d\theta$.
Using the standard integral result $\int_0^{\pi/2} \log \sin \theta \, d\theta = -\frac{\pi}{2} \log 2$,we get:
$I = \frac{2}{\pi} \left( -\frac{\pi}{2} \log 2 \right) = -\log 2$.

(C) Let $I = \int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx$.
Substitute $x = \sin \theta$,then $dx = \cos \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x = 1$,$\theta = \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int_0^{\pi/2} \frac{\log(\sin \theta)}{\sqrt{1 - \sin^2 \theta}} \cdot \cos \theta \, d\theta$
$I = \int_0^{\pi/2} \frac{\log(\sin \theta)}{\cos \theta} \cdot \cos \theta \, d\theta$
$I = \int_0^{\pi/2} \log(\sin \theta) \, d\theta$
Using the standard definite integral result $\int_0^{\pi/2} \log(\sin \theta) \, d\theta = - \frac{\pi}{2} \log 2$,we get:
$I = - \frac{\pi}{2} \log 2$.

(B) Let $I = \int_0^{\pi /2} {x\cot x\,dx}$.
Using integration by parts,let $u = x$ and $dv = \cot x \, dx$. Then $du = dx$ and $v = \log(\sin x)$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = [x \log(\sin x)]_0^{\pi /2} - \int_0^{\pi /2} \log(\sin x) \, dx$.
Evaluating the boundary term:
At $x = \pi/2$,$x \log(\sin x) = \frac{\pi}{2} \log(1) = 0$.
As $x \to 0^+$,$x \log(\sin x) \to 0$ (using $L$'Hopital's rule).
Thus,$[x \log(\sin x)]_0^{\pi /2} = 0$.
We know that $\int_0^{\pi /2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2$.
Substituting this back into the expression for $I$:
$I = 0 - (-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2$.

(B) Let $I = \int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$.
Substitute $t = x + 1$,so $dt = dx$.
When $x = -2$,$t = -1$. When $x = 0$,$t = 1$.
Also,$x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2 = t^3 + 2$.
Thus,$I = \int_{-1}^{1} [t^3 + 2 + t\cos(t)] \, dt$.
We can split this into $I = \int_{-1}^{1} t^3 \, dt + \int_{-1}^{1} 2 \, dt + \int_{-1}^{1} t\cos(t) \, dt$.
Since $t^3$ and $t\cos(t)$ are odd functions,their integrals over the symmetric interval $[-1, 1]$ are $0$.
Therefore,$I = 0 + \int_{-1}^{1} 2 \, dt + 0 = [2t]_{-1}^{1} = 2(1 - (-1)) = 2(2) = 4$.