Download App

STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
If $f(x)=\int_{0}^{x} t \sin t \,d t,$ then $f^{\prime}(x)$ is
  • A
    $\cos \,x+x \sin\, x$
  • ✓
    $x\, \sin \,x$
  • C
    $x\, \cos \,x$
  • D
    $\sin\, x+x \,\cos \,x$
✓

Answer

Correct option: B.
$x\, \sin \,x$
b
$f(x)=\int_{0}^{x} t \sin t d t$

Integrating by parts, we obtain

$f(x)=t \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \int \sin t \, d t\right\} d t$

$=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cot t) d t$

$=[-t \cos t+\sin t]_{0}^{x}$

$=-x \cos x+\sin x$

$\Rightarrow f^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x$

$=x \sin x-\cos x+\cos x$

$=x \sin x$

Hence, the correct Answer is $B$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{}^{} {x{{\cos }^2}} xdx = $
→
The value of $k$ for which the function $f\left( x \right) = \left\{ \begin{gathered} {\left( {\frac{4}{5}} \right)^{\frac{{\tan \,4x}}{{\tan \,5x}}}},\,\,\,\,0 < x < \frac{\pi }{2} \hfill \\  k + \frac{2}{5}\,\,\,,\,\,\,\,\,\,\,\,\,\,\,x = \frac{\pi }{2} \hfill \\ \end{gathered}  \right.$ is continuous at  $x\,= \frac{\pi}{2}$ is
→
Let $f(x + y) = f(x) + f(y)$and $f(x) = {x^2}g(x)$ for all $x,y \in R$, where $g(x)$ is continuous function. Then $f'(x)$ is equal to
→
In two events $P(A \cup B) = 5/6$, $P({A^c}) = 5/6$, $P(B) = 2/3,$ then $A$ and $B$ are
→
If $a = 2i + k,\,\,b = i + j + k$ and $c = 4i - 3j + 7k.$ If $d \times b = c \times b$ and $d\,.\,a = 0,$ then $d$ will be
→
If coordinates of the points $A$ and $ B$ are $(2, 4)$ and $(4, 2)$ respectively and point M is such that $A-M-B$ also $AB =\, 3\, AM$, then the coordinates of $ M$ are
→
If $n (2 n +1) \int_{0}^{1}\left(1- x ^{ n }\right)^{2 n } dx =1177 \int_{0}^{1}\left(1- x ^{ n }\right)^{2 n +1} dx$, then $n \in N$ is equal to $\dots\dots$
→
$\left|x+\frac{1}{x}\right| \geq 2$ then $x \in$
→
If $\cos x = {1 \over {\sqrt {1 + {t^2}} }}$ and $\sin y = {t \over {\sqrt {1 + {t^2}} }}$, then ${{dy} \over {dx}} = $
→
Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{x}^{2}-\mathrm{x}-1=0 .$ If $\mathrm{p}_{\mathrm{k}}=(\alpha)^{\mathrm{k}}+(\beta)^{\mathrm{k}}, \mathrm{k} \geq 1,$ then which one of the following statements is not true?
→