If the area bounded by the curve y = x^3 + ax | vedclass

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(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx = \frac{37}{4}$.Since $a > 0$, $x^3 + ax = x(x^2 + a) = 0$ at $x = 0$ because $x^2 + a >

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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 $a > 0$, $x^3 + ax = x(x^2 + a) = 0$ at $x = 0$ because $x^2 + a > 0$.For $x \in [-2, 0]$, $x^3 + ax \le 0$. For $x \in [0, 1]$, $x^3 + ax \ge 0$.Area $= \int_{-2}^{0} -(x^3 + ax) \, dx + \int_{0}^{1} (x^3 + ax) \, dx = \frac{37}{4}$.Evaluating the integrals: $-[\frac{x^4}{4} + \frac{ax^2}{2}]_{-2}^{0} + [\frac{x^4}{4} + \frac{ax^2}{2}]_{0}^{1} = \frac{37}{4}$.$-[0 - (\frac{16}{4} + \frac{4a}{2})] + [(\frac{1}{4} + \frac{a}{2}) - 0] = \frac{37}{4}$.$4 + 2a + \frac{1}{4} + \frac{a}{2} = \frac{37}{4}$.$\frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = \frac{20}{4} = 5$.$a = 5 	imes \frac{2}{5} = 2$.

If the area bounded by the curve $y = x^3 + ax$ (where $a > 0$), the $X$-axis, and the lines $x = -2$ and $x = 1$ is $\frac{37}{4} \text{ square units}$, find the value of $a$.

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