If the area bounded by y = x^3 + ax (where a | vedclass

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(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx = \frac{37}{4}$.Since $x^3 + ax = x(x^2 + a)$, the function is negative for $x \in [-2, 0

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$a = 2$

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(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx = \frac{37}{4}$.Since $x^3 + ax = x(x^2 + a)$, the function is negative for $x \in [-2, 0]$ and positive for $x \in [0, 1]$ (given $a > 0$).Thus, $A = \int_{-2}^{0} -(x^3 + ax) \, dx + \int_{0}^{1} (x^3 + ax) \, dx = \frac{37}{4}$.Evaluating the integrals:$-\left[ \frac{x^4}{4} + \frac{ax^2}{2} \right]_{-2}^{0} + \left[ \frac{x^4}{4} + \frac{ax^2}{2} \right]_{0}^{1} = \frac{37}{4}$.$-(0 - (4 + 2a)) + (\frac{1}{4} + \frac{a}{2}) = \frac{37}{4}$.$4 + 2a + \frac{1}{4} + \frac{a}{2} = \frac{37}{4}$.$\frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = \frac{20}{4} = 5$.$a = 2$.

If the area bounded by $y = x^3 + ax$ (where $a > 0$), the $x$-axis and the lines $x = -2$ and $x = 1$ is $\frac{37}{4}$ square units, then:

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