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 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

Answer: Option D. -> loge 2
:
D
${lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}]$
$={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}]$
$=\frac{1}{n}{lim}_{n\to \infty }[1+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots +\frac{1}{1+\frac{n}{n}}]$
$=\frac{1}{n}{lim}_{n\to \infty }{\sum }_{r=0}^n\left[\frac{1}{1+\frac{r}{n}}\right]={\int }_0^1\frac{1}{1+x}dx$
$=\left[lo{g}_e\right(1+x){]}_0^1=lo{g}_{e}2-lo{g}_{e}1=lo{g}_{e}2$

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 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. -> 15
:
C
We have seen that if $g\left(y\right)\ge 0$ for $yϵ$ [c, d] then the area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is ${\int }_{(y=c)}^{d}g\left(y\right)dy$ .
We'll use the same concept here. Here the curve given is
$x=y^2+2$ and c = 0 $\&d=3.So,g\left(y\right)\ge 0\forall xϵ(0,3)$
Let the area enclosed be A.
$A={\int }_0^3(y^2+2)dy.$
$A=(\frac{y^3}{3}+2y){|}_0^{3}A=9+6-0A=15$

Answer: Option A. -> 43
:
A
$y^2=xand2y=x\Rightarrow y^2=2y\Rightarrow y=0,2$
$\therefore Requiredarea={\int }_0^2(y^2-2y)dy={(\frac{y^3}{3}-y^2)}_0^2=\frac{4}{3}sq.unit$

Answer: Option A. -> True
:
A
With all the information given let’s draw the graph of $f\left(x\right)$ 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.
$\underset{n\to \infty }{lim}\frac{b-a}{n}$ 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, $\underset{n\to \infty }{lim}{s}_n$ = Area PRSQP = $\underset{n\to \infty }{lim}{S}_n$
$\frac{b-a}{n}=h\left(given\right)$
So, Area of the graph =
$\underset{n\to \infty }{lim}\frac{b-a}{n}\left[f\right(a)+f(a+h)...f(a+(n-1)h\left)\right]$

Answer: Option D. -> Hyperbola
:
D
$I_{n+1}={\int }_0^{\frac{\pi }{2}}{\cos }^{n+1}x\cos (n+1)xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n+1}x(\cos nx\cos x-\sin nx\sin x)dx\therefore I_{n+1}=I_n-I_{n+1}\Rightarrow 2I_{n+1}=I_n\therefore I_n:I_{n+1}=2$

Answer: Option B. -> 2√2−2
:
B
$y={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{\sqrt{n^2+n}}+\cdots +\frac{1}{\sqrt{n^2+(n-1)n}}]$
$\Rightarrow y={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n\sqrt{1+\frac{1}{n}}}+\cdots +\frac{1}{n\sqrt{1+\frac{(n-1)}{n}}}]$
$\Rightarrow y=\frac{1}{n}{lim}_{n\to \infty }[1+\frac{1}{\sqrt{1+\frac{1}{n}}}+\cdots +\frac{1}{\sqrt{1+\frac{(n-1)}{n}}}]$
$y={lim}_{n\to \infty }\frac{1}{n}{\sum }_n^{k=1}\frac{1}{\sqrt{1+\frac{(k-1)}{n}}},Put\frac{k-1}{n}=xand\frac{1}{n}=dx$
$\Rightarrow y={lim}_{n\to \infty }{\int }_0^{\frac{n-1}{n}}\frac{dx}{\sqrt{1+x}}={lim}_{n\to \infty }2{\left[\sqrt{1+x}\right]}_0^{\left(\frac{n-1}{n}\right)}$
$\Rightarrow y=2{lim}_{n\to \infty }[\sqrt{\frac{2n-1}{n}}-1]=2{lim}_{n\to \infty }\sqrt{\frac{2n-1}{n}}-2$
$\Rightarrow y=2{lim}_{n\to \infty }\sqrt{2-\frac{1}{n}}-2=2\sqrt{2}-2$