MCQ
$I = \int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is equal to
  • A
    $4$
  • $0$
  • C
    ${2^{1/3}}$
  • D
    None of these

Answer

Correct option: B.
$0$
b
Apply king's rule $I = -I$

Hence $I = 0$

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

If the fucnction $\text{f(x)}=\begin{cases}(\cos\text{x})^{\frac{1}{\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuouse at x = 0, then the value of k is:
Find the degree of the differential equation:
$\Big(1+\frac{\text{dx}}{\text{dy}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Statement $1$ : A function $f:R \to R$ is continuous at $x_0$ if and only if $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right)$ exists and $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right) = f\left( {{x_0}} \right)$

Statement $2$ :  A function $f : R \to R$ is discontinuous at $x_0$ if and only if, $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right)$ exists and $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right) \ne f\left( {{x_0}} \right)$

Let $d$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1} \quad$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $\mathrm{d}^2+6$ is equal to :
The solution of the differential equation $ydx - \left( {x + 2{y^2}} \right)dy = 0$ is $x\, = f(y)$. If $f(-1)\, = 1$, then $f(1)$ is equal to
Let $I_n=\int \limits_1^0(\log x)^n d x$, where $n$ is a non-negative integer. Then, $I_{2011}+2011\,\,\,\,\,I_{2010}$ is equal to
The general solution of the differential equation, $x \left( {\frac{{dy}}{{dx}}} \right) = y \cdot \ln \left( {\frac{y}{x}} \right)$ is :
where $c$ is an arbitrary constant.
Choose the correct answer in Exercise:$\int^\frac{2}{3}_{0}\frac{\text{dx}}{4+9\text{x}^{2}}\text{equals}$
$\int_1^e {\frac{{1 + \log x}}{x}\,dx = } $
If $\text{f(x)}=\begin{cases}\frac{1-\cos10\text{x}}{\text{x}^2},&\text{x}<0\\\text{a},&\text{x}=0\\\frac{\sqrt{\text{x}}}{\sqrt{625+\sqrt{\text{x}}}-25},&\text{x}>0\end{cases}$ then the value of so that f(x) may be continuous at x = 0 is: