Download App

7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} = $
  • ✓
    $ - \frac{1}{2}{e^{ - \pi /2}}$
  • B
    $ - \frac{{\sqrt 2 }}{2}{e^{ - \pi /4}}$
  • C
    $ - \sqrt 2 ({e^{ - \pi /4}} + {e^{ - \pi /4}})$
  • D
    $0$
✓

Answer

Correct option: A.
$ - \frac{1}{2}{e^{ - \pi /2}}$
a
(a) $\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} $

$= \left[ {\frac{{{e^{ - x}}}}{2}( - \sin x - \cos x)} \right]_{ - \pi /4}^{\pi /2}$

$ = \frac{1}{2}[{e^{ - x}}( - \sin x - \cos x)]_{ - \pi /4}^{\pi /2}$

$ = \frac{1}{2}\left[ {{e^{ - \pi /2}}( - 1 - 0) - \left\{ {{e^{\pi /4}}\left( {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}} \right)} \right\}} \right] $

$= - \frac{{{e^{ - \pi /2}}}}{2}$.

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

Given that  $ a$ and $  b$ are two unit non-collinear vectors, if $u = a - (a\,.\,b)\,b$ and $v = a \times b$, then find $|v| =$.
→
Evaluate: $\int\left(2^x+2^{-x}\right)^2 d x$
→
Choose the correct answer from the given four options.
Let f : R → R be given by f(x) = tanx. Then f-1(1) is:
  1. $\frac{\pi}{4}$
  2. $\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\}$
  3. $\text{Does not exist.}$
  4. $\text{None of these}.$
→
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
  1. $-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
  2. $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
  3. $-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
  4. $2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
→
Consider a rectangle whose length is increasing at the uniform rate of $2\, m/sec$, breadth is decreasing at the uniform rate of $3\, m/sec$ and the area is decreasing at the uniform rate of $5\,m^2/ sec$ . If after some time the breadth of the rectangle is $2\, m$ then the length of the rectangle is ........ $m.$
→
The area of figure bounded by $y = {e^x},\,y = {e^{ - x}}$ and the straight line $x = 1$ is
→
If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a}>\text{b}>0,$then:
  1. $\text{y}_1=\frac{-1}{\text{a}+\text{b}\cos\text{x}}$
  2. $\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
  3. $\text{y}_1=\frac{1}{\text{a}-\text{b}\cos\text{x}}$
  4. $\text{y}_2=\frac{-\text{b}\sin\text{x}}{(\text{a}-\text{b}\cos\text{x})^2}$
→
The derivative of $\cos^{-1}(2\text{x}^2-1)$ with respect to $\cos^{-1}\text{x}$ is:
  1. $2$
  2. $\frac{1}{2\sqrt{1+\text{x}^2}}$
  3. $\frac{2}{\text{x}}$
  4. $1-\text{x}^2$
→
If ${\tan ^{ - 1}}\frac{{a + x}}{a} + {\tan ^{ - 1}}\frac{{a - x}}{a} = \frac{\pi }{6}$,then ${x^2} =$
→
Let $\vec p\, = \,2\hat i\, + \,3\hat j\, + \,a\hat k$, $\vec q\, = \,b\hat i\, + \,5\hat j\, - \hat k$, $\vec r\, = \,\hat i\, + \,\hat j\, + 3\hat k$ . If $\vec p,\vec q,\vec r$ are coplanar and $\vec p,\vec q$= $20$ , then the ordered pair $(a, b)$ is 
→