If $\int_a^b x^3 d x=0$ and $\int_a^b x^2 d x=\frac{2}{3}$, then

If $\int_a^b x^3 d x=0$ and $\int_a^b x^2 d x=\frac{2}{3}$, then
  1. $a=-1$ and $b=1$
  2. $a=1$ and $b=-1$
  3. $a=2$ and $b=-2$
  4. $a=-2$ and $b=2$

Solution

We have, $ \begin{aligned} & \int_a^b x^3 d x=0 \\ & \Rightarrow \quad\left[\frac{x^4}{4}\right]_a^b=0 \Rightarrow b^4-a^4=0 \\ & \Rightarrow \quad b=-a, b \neq a \\ & \text { and } \quad \int_a^b x^2 d x=\frac{2}{3} \\ & \Rightarrow \quad\left[\frac{x^3}{3}\right]_a^b=\frac{2}{3} \\ & \Rightarrow \quad b^3-a^3=2 \\ & \Rightarrow \quad-a^3-a^3=2 \\ & \end{aligned} $ $\begin{array}{ll}\Rightarrow & a^3=-1 \\ \Rightarrow & a=-1 \\ \therefore & b=-a=1\end{array}$

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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