Answer: Option D. -> 3π8
:
D
${\int }_0^{\pi }|si{n}^{4}x|dx=2{\int }_0^{\frac{\pi }{2}}si{n}^{4}xdx$
Applying gamma function,
$2{\int }_0^{\frac{\pi }{2}}si{n}^{4}xdx=2\frac{T\left(\frac{5}{2}\right).T\left(\frac{1}{2}\right)}{2.T\left(\frac{6}{2}\right)}=\frac{3\pi }{8}$

Answer: Option C. -> 812
:
C
The given equation of curve can be written as $x=f\left(y\right)=9y-y^3$.Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
$f\left(y\right)=y(9-y^2)$
$f\left(y\right)=y.(3+y)(3-y)$
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
$A={\int }_{-3}^0|f\left(y\right)|dy+{\int }_0^{3}f\left(y\right)dyA={\int }_{-3}^0|9y-y^3|dy+{\int }_0^3(9y-y^3)dyA=-{\int }_{-3}^0(9y-y^3)dy+{\int }_0^{3}9y-y^{3}dyA=-[\frac{9y^2}{2}-\frac{y^4}{4}{]}_{-3}^0+[\frac{9y^2}{2}-\frac{y^4}{4}{]}_0^{3}A=-[0-0-\frac{81}{2}+\frac{81}{4}]+[\frac{81}{2}-\frac{81}{4}]A=81-\frac{81}{2}A=\frac{81}{2}$

Answer: Option A. -> True
:
A
With all the information given let’s draw the graph of f(x) on cartesian coordinate system.

The area which we are interested in calculating is the area PRSQP. To calculate this area let’s divide this into rectangles of equal and infinitesimally small width. Now, what’s the purpose of this? Why are we doing this? Let’s discuss them one by one.
First why rectangle? Very simple, we do not know any approach to calculate the area of graphs like this. But we do know how to calculate the area of rectangle.
Second why infinitesimally small width? Just because we know how to calculate the area of rectangle we can’t calculate the area of this graph.
For example if we divide the stretch a to b in two parts, i.e. dividing it into two rectangles of equal width will give us area which is not equal to the area of this graph. The difference between the area of the rectangles and the area of this graph will be significant. So, to reduce this error of the difference we’ll divide this graph into infinite rectangles.
So, the the length a to b which is b -a will be divided into n equal parts giving width as $\frac{b-a}{n}$
Now consider regions ABCD, ABLC, ABDM.
By observation we can say -
Area of the rectangle ABLC < area of the region ABCD < Area of the rectangle ABDM
Actually when the width becomes infinitesimally small all these area become nearly equal to each other.
lim $n\to \infty \frac{b-a}{n}$ Area of the rectangle ABLC $\approx$ Area of the region ABCD $\approx$ Area of the rectangle ABDM
Now we form the sums -
$s_n=\frac{b-a}{n}\left[f\right(a)+f(a+h)...f(a+(n-1)h\left)\right]$
$S_n=\frac{b-a}{n}\left[f\right(a+h)+f(a+2h)...f(a+nh\left)\right]$
Where, $s_n\&S_n$ denote the sum of the areas of lower rectangles and upper rectangles respectively.
And we know, $s_n<AreaPRSQP<S_n$
As $n\to \infty$ strips becomes narrower and narrower and both the sum becomes equal.
So, lim $n\to \infty s_n$ = Area PRSQP = lim $n\to \infty S_n$
$\frac{b-a}{n}=h\left(given\right)$
So, Area of the graph =
lim $n\to \infty \frac{b-a}{n}\left[f\right(a)+f(a+h)...f(a+(n-1)h\left)\right]$

Answer: Option B. -> x3ex
:
B
$f\left(x\right)={\int }_a^x{t}^3{e}^{t}dt={\int }_a^0{t}^3.e^{t}dt+{\int }_0^x{t}^3{e}^{t}dt$
$\Rightarrow \frac{df\left(x\right)}{dx}=\frac{d}{dx}\left({\int }_a^0{t}^3.e^{t}dt\right)+\frac{d}{dx}\left({\int }_0^x{t}^3.e^{t}dt\right)=x^3{e}^x$

Answer: Option A. -> 2
:
A
The two curves $y^2$ = 4ax and y = mx intersect at $(\frac{4a}{m^2},\frac{4a}{m})$ and the area enclosed by the two curves is given by ${\int }_0^{\frac{4a}{m^2}}(\sqrt{4ax}-mx)dx$
$\therefore {\int }_0^{\frac{4a}{m^2}}(\sqrt{4ax}-mx)dx=\frac{a^2}{3}$
$\Rightarrow \frac{8}{3}\frac{a^2}{m^3}=\frac{a^2}{3}\Rightarrow m^3=8\Rightarrow m=2$

Answer: Option A. -> log 2
:
A
put 1+ sin x =t
$Then{\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+sinx}dx={[log|1+sinx|]}_0^{\frac{\pi }{2}}=log2$

Answer: Option B. -> 13
:
B
$I={\int }_0^1\frac{x^7}{\sqrt{1-x^4}}dx={\int }_0^1\frac{x^{6}xdx}{\sqrt{1-x^4}}$
$Put{x}^2=sin\theta \Rightarrow 2xdx=cos\theta d\theta$
$I=\frac{1}{2}{}_0^{\frac{\pi }{2}}\frac{si{n}^3\theta .cos\theta d\theta }{cos\theta }=\frac{1}{2}{\int }_0^{\frac{1}{2}}si{n}^3\theta d\theta$
$=\frac{1}{2}\frac{T2T\left(\frac{1}{2}\right)}{2.T\left(\frac{1}{2}\right)}=\frac{T\left(\frac{1}{2}\right)}{4.\frac{3}{2}.\frac{1}{2}.T\left(\frac{1}{2}\right)}=\frac{1}{3}$

Answer: Option C. -> 12
:
C
$I_1={\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx$
$={\int }_{1-k}^k(1-k+k-x)f\left[\right(1-k+k-x\left)\right\{1-(1-k+k-x)\left\}\right]dx$
$={\int }_{1-k}^k(1-x)f\left\{x\right(1-x\left)\right\}dx$
$={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx-{\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx=I_2-I_1$
$\therefore 2I_1=I_2\Rightarrow \frac{I_1}{I_2}=\frac{1}{2}$

Answer: Option A. -> π log 2
:
A
${\int }_0^{\frac{\pi }{2}}log(tanx+cotx)dx={\int }_0^{\frac{\pi }{2}}log\left[\frac{2}{sin2x}\right]dx$
$={\int }_0^{\frac{\pi }{2}}(log2-logsin2x)dx=log2\left(\frac{\pi }{2}\right)+\left(\frac{\pi }{2}\right)log2=\pi log2$

Answer: Option D. -> I=I4
:
D
For 0 < x < 1, we have $\frac{1}{2}{x}^2<x^2<x$
$\Rightarrow -x^2>-x,sothat{e}^{-x^2}<e^{-x},$
$Hence{\int }_0^1{e}^{-x^2}co{s}^{2}xdx>{\int }_0^1{e}^{-x}co{s}^{2}xdx$
$Alsoco{s}^{2}x\le 1$
$Therefore{\int }_0^1{e}^{-x^2}co{s}^{2}xdx\le {\int }_0^1{e}^{-x^2}dx<{\int }_0^1{e}^{\frac{-x^2}{2}}dx=I_4$
Hence $I_4$ is the greatest integral