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

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(D) Given $\int_a^b {x^3} dx = 0$.Evaluating the integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \implies b^4 - a^4 = 0 \implies (b^2 - a^2)(b^2 + a^2

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$-1, 1$

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(D) Given $\int_a^b {x^3} dx = 0$.Evaluating the integral: $\left[ \frac{x^4}{4} \right]_a^b = 0 \implies b^4 - a^4 = 0 \implies (b^2 - a^2)(b^2 + a^2) = 0$.Since $a$ and $b$ are real,$b^2 + a^2 = 0$ implies $a=0, b=0$,which contradicts the second integral. Thus,$b^2 - a^2 = 0$,which means $b = a$ or $b = -a$.If $b = a$,the integral $\int_a^b {x^2} dx = 0$,but we are given $\int_a^b {x^2} dx = \frac{2}{3}$. Thus,$b = -a$.Given $\int_a^b {x^2} dx = \frac{2}{3}$,we have $\left[ \frac{x^3}{3} \right]_a^b = \frac{2}{3} \implies b^3 - a^3 = 2$.Substituting $b = -a$: $(-a)^3 - a^3 = 2 \implies -2a^3 = 2 \implies a^3 = -1 \implies a = -1$.Since $b = -a$,$b = -(-1) = 1$.Therefore,$(a, b) = (-1, 1)$.

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:

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