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: Understanding the Concept:
Since \(a > 0\), the function \(y = x^3 + ax = x(x^2 + a)\) is negative for \(x < 0\) and positive for \(x > 0\). It crosses the axis only at \(x = 0\). So area must be found in two pieces.

Step 2: Key Formula or Approach:
\[ \text{Area} = \int_{-2}^{0}\left|x^3 + ax\right|dx + \int_0^1\left(x^3 + ax\right)dx \]

Step 3: Detailed Explanation:
Left piece, with the sign reversed:
\[ -\int_{-2}^{0}(x^3 + ax)\,dx = -\left[\frac{x^4}{4} + \frac{ax^2}{2}\right]_{-2}^{0} = \frac{16}{4} + \frac{4a}{2} = 4 + 2a \]
Right piece:
\[ \int_0^1(x^3 + ax)\,dx = \frac14 + \frac{a}{2} \]
Total area:
\[ 4 + 2a + \frac14 + \frac a2 = \frac{17}{4} + \frac{5a}{2} \]
Set this to \(\dfrac{37}{4}\):
\[ \frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = 5 \Rightarrow a = 2 \]
Option (A) \(a = 4\) would give area \(\tfrac{17}{4} + 10\), too large. Options (C) and (D) are even larger.

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