Download App

7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) d x}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}$ is equal to
  • A
    $1$
  • B
    $2$
  • C
    $4$
  • ✓
    $3$
✓

Answer

Correct option: D.
$3$
d
$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right)}{\left(x^{2}+2\right) \sqrt{4+x^{4}}} d x$

$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{x^{2}\left(\frac{2}{x^{2}}-1\right) d x}{x\left(x+\frac{2}{x}\right) \times x \sqrt{\frac{4}{x^{2}}+x^{2}}}$

$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(\frac{2}{x^{2}}-1\right) d x}{\left(x+\frac{2}{x}\right) \sqrt{\left(x+\frac{2}{x}\right)^{2}-4}}$

$x +\frac{2}{ x }= t$

$dt =\left(1-\frac{2}{ x ^{2}}\right) dx$

$I =-\frac{24}{\pi} \int \frac{ dt }{ t \sqrt{ t ^{2}-4}}$

$=-\frac{24}{\pi} \times \frac{1}{2} \sec ^{-1}\left[\frac{ x +\frac{2}{ x }}{2}\right)^{\sqrt{2}}$

$=-\frac{12}{\pi}\left[\sec ^{-1}\left(\frac{2 \sqrt{2}}{2}\right)-\sec ^{-1}(\infty)\right]$

$=-\frac{12}{\pi}\left[\frac{\pi}{4}-\frac{2 \pi}{2 \times 2}\right]=-\frac{12}{\pi}\left[-\frac{\pi}{4}\right]$

$=3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 7.2 definite integral→7.2 definite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $f(x) = \frac{{{{\cos }^2}x + {{\sin }^4}x}}{{{{\sin }^2}x + {{\cos }^4}x}}$ for $x \in R$, then $f(2002) = $
→
If $ab = 2a + 3b,\,a > 0,\,\,b > 0$ then the minimum value of  $ ab$  is
→
If $\int {\frac{{2{x^2} + 3.dx}}{{({x^2} - 1)({x^2} - 4)}}} = \log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^a}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^b} + c$ then the values of $a$ and $b$ respectively are
→
Let $\quad \overrightarrow{ a }=2 \hat{ i }-7 \hat{ j }+5 \hat{ k } \quad, \quad \overrightarrow{ b }=\hat{ i }+\hat{ k } \quad$ and $\overrightarrow{ c }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$ be three given vectors. If $\overrightarrow{ r }$ is a vector such that $\overrightarrow{ r } \times \overrightarrow{ a }=\overrightarrow{ c } \times \overrightarrow{ a }$ and $\overrightarrow{ r } \cdot \overrightarrow{ b }=0$, then $|\overrightarrow{ r }|$ is equal to :
→
Let $A$ and $B$ be two $3 \times 3$ matrices such that $AB = I$ and $| A |=\frac{1}{8}$ then $|\operatorname{adj}( Badj (2 A ))|$ is equal to
→
The shortest distance between line $y -x = 1$ and curve $x = y^2$ is
→
Which of the following differentials equation has y = x as one of its particular solution?
  1. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
  2. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
  3. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
  4. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
→
Given that $f (x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f (a) < 0$ and $f (b) > 0$, which of the following are always true ?

$(i)$ $f (x)$ is bounded on $a \le x \le b.$

$(ii)$ The equation $f (x) = 0$ has at least one solution in $a < x < b.$

$(iii)$ The maximum and minimum values of $f (x)$ on $a \le x \le b$ occur at points where $f ' (c) = 0$.

$(iv)$ There is at least one point $c$ with $a < c < b$ where $f ' (c) > 0$.

$(v)$ There is at least one point $d$ with $a < d < b$ where $f ' (c) < 0.$

→
If $x$  takes non-positive permissible value, then ${\sin ^{ - 1}}x =$
→
The objective function of LPP defined over the convex set attains its optimum value at.
→