Question

If ${\int }_a^b{x}^{3}dx=0$ and ${\int }_a^b{x}^{2}dx=\frac{2}{3}$, then the values of $a$ and $b$ are respectively,

• A. $1,1$
• B. $-1,-1$
• C. $1,-1$
• D. $-1,1$

Answer 1

The correct option is D

$-1,1$

Explanation for the correct option.

Step 1. Find the first condition.

The integral equation ${\int }_a^b{x}^{3}dx=0$ can be simplified as:

--

So, $b^2+a^2=0$ is not possible, thus $b^2-a^2=0$ and so $b=\pm a$.

Step 2. Find the values of $a$ and $b$.

The integral equation ${\int }_a^b{x}^{2}dx=\frac{2}{3}$ can be simplified as:

$$\begin{gathered} {\int }_a^b{x}^{2}dx=\frac{2}{3} \\ \Rightarrow {\left[\frac{x^3}{3}\right]}_a^b=\frac{2}{3} \\ \Rightarrow \frac{b^3}{3}-\frac{a^3}{3}=2 \\ \Rightarrow b^3-a^3=2 \end{gathered}$$

Now, case- $1$: $b=a$ then

$$\begin{gathered} a^3-a^3=2 \\ \Rightarrow 0=2 \end{gathered}$$

This produces a false statement and so the case is rejected.

Again, case- $2$: $b=-a$ then

$$\begin{gathered} {(-a)}^3-a^3=2 \\ \Rightarrow -2a^3=2 \\ \Rightarrow a^3=-1 \\ \Rightarrow a=-1 \end{gathered}$$

And as $b=-a$, so $b=1$.

So the values of $a$ and $b$ are respectively, $-1$ and $1$.

Hence, the correct option is D.