Number of ordered pairs (a, b) satisfying sim | vedclass

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(B) Given the system of equations:$1$) $\int_{a}^{b} x^{3} dx = 0$$2$) $\int_{a}^{b} x^{2} dx = \frac{2}{3}$Solving the first equation:$\left[\frac{x^

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(B) Given the system of equations:$1$) $\int_{a}^{b} x^{3} dx = 0$$2$) $\int_{a}^{b} x^{2} dx = \frac{2}{3}$Solving the first equation:$\left[\frac{x^{4}}{4}\right]_{a}^{b} = 0 \Rightarrow \frac{b^{4}-a^{4}}{4} = 0 \Rightarrow b^{4} = a^{4}$This implies $b = a$ or $b = -a$.Case $1$: If $b = a$,then $\int_{a}^{a} x^{2} dx = 0$,which contradicts $\int_{a}^{b} x^{2} dx = \frac{2}{3}$.Case $2$: If $b = -a$,substitute into the second equation:$\int_{a}^{-a} x^{2} dx = \frac{2}{3}$$\left[\frac{x^{3}}{3}\right]_{a}^{-a} = \frac{2}{3}$$\frac{(-a)^{3} - a^{3}}{3} = \frac{2}{3}$$\frac{-a^{3} - a^{3}}{3} = \frac{2}{3}$$-2a^{3} = 2 \Rightarrow a^{3} = -1 \Rightarrow a = -1$Since $b = -a$,we get $b = 1$.Thus,there is only one ordered pair $(a, b) = (-1, 1)$ that satisfies the system.

Number of ordered pairs $(a, b)$ satisfying simultaneously the system of equations $\int_{a}^{b} x^{3} dx = 0$ and $\int_{a}^{b} x^{2} dx = \frac{2}{3}$ is

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