Answer: Option D. -> −√2−√3+5
:
D
${\int }_0^2\left[x^2\right]dx$
$={\int }_0^1\left[x^2\right]dx+{\int }_0^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right]dx+{\int }_{\sqrt{3}}^2\left[x^2\right]dx$
$={\int }_0^{1}0dx+{\int }_0^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$
$=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}$
$=5-\sqrt{3}-\sqrt{2}$

Answer: Option A. -> 20
:
A
$\therefore {\int }_0^{10\pi }|sinx|dx=10{\int }_0^{\pi }|sinx|dx$
$\therefore |sinx|is$ positive in I & II quadrant and has period $\pi$
$=10{\int }_0^{\pi }sinxdx=10[-cosx{]}_0^x=20$

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. -> False
:
A
Let’s calculate its value.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll change the limit which is not finite to a variable and then put the limits.
${\int }_1^{\infty }\frac{1}{x^2}dx={lim}_{a\to \infty }{\int }_1^a\frac{1}{x^2}dx$
= lim $a\to \infty (-\frac{1}{x}){|}_1^a$ Since, $\int \frac{1}{x^2}dx=(-\frac{1}{x})$
$=0-(-1)=1$

Answer: Option C. -> 12  
:
C
Let $I={\int }_0^{\frac{1}{3}}0dx+{\int }_{\frac{1}{3}}^{\frac{2}{3}}1dx+{\int }_{\frac{2}{3}}^{1}2dx+{\int }_1^{\frac{4}{3}}3dx+{\int }_{\frac{4}{3}}^{\frac{5}{3}}4dx+{\int }_{\frac{5}{3}}^{\frac{6}{3}}5dx+{\int }_{\frac{6}{3}}^{\frac{7}{3}}6dx+{\int }_{\frac{7}{3}}^{\frac{8}{3}}7dx+{\int }_{\frac{8}{3}}^{\frac{9}{3}}8dx=\frac{1}{3}(1+2+3+4+5+6+7+8)=12$

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 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. -> False
:
A
We have seen that there are some functions, antiderivatives of which cannot be written in terms of elementary functions. Functions like $e^{x^2},\frac{e^{x^2}}{x},sin\left(\frac{1}{x}\right),$etc. are some of the examples of such functions.

Answer: Option A. -> False
:
A
We have seen that there are functions which have discontinuity but still can be integrated. These functions have finite type discontinuities. Please try to understand one thing here, that definite integral gives you the area so area can be calculated by the curve into pieces as well.
For example -

We can see that the given function is discontinuous at x =1. So to integrate this function from (-3,3) we’ll have to integrate it into pieces. One piece will be (-3,1) and another will be (1, 3).
${\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^{1}f\left(x\right)dx+{\int }_1^{3}f\left(x\right)dx$
And given that for $(-3,1)f\left(x\right)=x^2$
And for (1,3) f(x) = 6-x
${\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^1\left(x^2\right)dx+{\int }_1^3(6-x)dx$
This way we can calculate the area even if the function is discontinuous.

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