Question

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 \(\ldots\)

• A. \(a = 4\)
• B. \(a = 2\)
• C. \(a = 10\)
• D. \(a = 20\)

Hint:

The curve is below the axis for x<0 and above for x>0. Split the area at x = 0.

Answer 1

The Correct Option is B

Step 1: Use the odd-function structure:
$y = x^3 + ax$ is odd. So the area from $-1$ to $0$ equals the area from $0$ to $1$, which is $\frac14 + \frac a2$.

Step 2: Add the extra strip:
The region from $-2$ to $-1$ has no mirror image inside the limits. By the same symmetry its area equals $\int_1^2 (x^3+ax)\,dx = \frac{16-1}{4} + a\cdot\frac{4-1}{2} = \frac{15}{4} + \frac{3a}{2}$.

Step 3: Total:
Area $= \left(\frac14 + \frac a2\right) + \left(\frac14 + \frac a2\right) + \left(\frac{15}{4} + \frac{3a}{2}\right) = \frac{17}{4} + \frac{5a}{2}$.

Step 4: Solve:
$\frac{17}{4} + \frac{5a}{2} = \frac{37}{4}$ gives $a = 2$.

Final Answer:
Option (B). \[ \boxed{a=2 \text{ (B)}} \]