If int_a^b x^3 dx = 0 and int_a^b x^2 dx = fr | vedclass

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(D) Given: $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.Step $1$: Evaluate the first integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \imp

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$-1, 1$

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(D) Given: $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.Step $1$: Evaluate the first integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \implies \frac{b^4 - a^4}{4} = 0 \implies b^4 = a^4 \implies b^2 = a^2$ or $b = -a$.Step $2$: Evaluate the second integral: $\left[ \frac{x^3}{3} \right]_a^b = \frac{2}{3} \implies \frac{b^3 - a^3}{3} = \frac{2}{3} \implies b^3 - a^3 = 2$.Step $3$: Substitute $b = -a$ into the second equation: $(-a)^3 - a^3 = 2 \implies -a^3 - a^3 = 2 \implies -2a^3 = 2 \implies a^3 = -1 \implies a = -1$.Step $4$: Since $b = -a$,we get $b = -(-1) = 1$.Thus,$a = -1$ and $b = 1$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

If $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$,then $a$ and $b$ are respectively

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