MCQ
$\int\limits_0^\pi  {} $ $(x · sin^2x · cos x) dx $ =
  • A
    $0$
  • B
    $2/9$
  • C
    $- 2/9$
  • $- 4/9$

Answer

Correct option: D.
$- 4/9$
d
$\int\limits_0^\pi  {x\,\cdot\,({{\sin }^2}x\;\cos x)\,dx} $= $x\,\cdot\,\left. {\frac{{{{\sin }^3}x}}{3}} \right|_{\,0}^{\,\pi } - \int\limits_0^\pi  {\frac{1}{3}{{\sin }^3}x\;dx} $

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

Similar questions

Let $f:[0, \infty) \rightarrow R$ be a continuous function such that $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ for all $x \in[0, \infty)$. Then, which of the following statement (s) is (are) $TRUE$?

$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$

$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$

$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$

$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$

Find the principal values of: $\sec ^{-1}(2)$
The set of value(s) of $ 'a'$ for which the function $f (x) = \frac{{a\,{x^3}}}{3} + (a + 2) x^2 + (a - 1) x + 2$ possess a negative point of inflection .
The general solution $y(x)$ of the differential equation $\frac{{d\left( {\int\limits_x^y {dt} } \right)}}{{dy}} = x$ , is
Let $\lambda $ be a real number for which the system of linear equations $x + y + z = 6$
 ; $4x + \lambda y - \lambda z = \lambda - 2$ ; $3x + 2y -4z = -5$ Has indefinitely many solutions. Then $\lambda $ is a root of the quadratic equation
If $c$ is any arbitrary constant, then the general solution of the differential equation $ydx - xdy = xy\,dx$ is given by
$\int_{}^{} {\frac{{dx}}{{{e^{ - 2x}}{{({e^{2x}} + 1)}^2}}} = } $
Choose the correct answer in the following:
The area bounded by the y-axis, y = cos x and y = sin x when $0\leq\text{x}\leq\frac{\pi}{2}$
Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 00$, and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1^{+}} g(x)=p$, then
If $A = \left( {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}{ - 5}&7&1\\1&{ - 5}&7\\7&1&{ - 5}\end{array}} \right)$ then $AB$ is equal to