Question

The value of \(\int\limits_{-3}^2x^2 |2x| dx\) is:

• A. 65
• B. \(-\frac{65}{2}\)
• C. 97
• D. \(\frac{97}{2}\)

Answer 1

The Correct Option is D

To evaluate the integral \(\int\limits_{-3}^2 x^2 |2x| \,dx\), we first consider the behavior of the absolute value function \(|2x|\).

The absolute value function is defined as:

\[ |2x| = \begin{cases} 2x, & x \geq 0 \\ -2x, & x < 0 \end{cases} \]

Since the function changes at \(x = 0\), we split the integral at 0:

\[ \int\limits_{-3}^{2} x^2 |2x| \,dx = \int\limits_{-3}^{0} x^2 (-2x) \,dx + \int\limits_{0}^{2} x^2 (2x) \,dx \]

1. For the first part:

\[ \int\limits_{-3}^0 x^2 (-2x) \,dx = -2 \int\limits_{-3}^0 x^3 \,dx \]

- The antiderivative of \(x^3\) is \(\dfrac{x^4}{4}\)

\[ -2 \left[ \frac{x^4}{4} \right]_{-3}^0 = -2 \left( \frac{0^4}{4} - \frac{(-3)^4}{4} \right) = -2 \left( 0 - \frac{81}{4} \right) = \frac{162}{4} = \frac{81}{2} \]

2. For the second part:

\[ \int\limits_0^2 x^2 (2x) \,dx = 2 \int\limits_0^2 x^3 \,dx \]

- The antiderivative of \(x^3\) is again \(\dfrac{x^4}{4}\)

\[ 2 \left[ \frac{x^4}{4} \right]_0^2 = 2 \left( \frac{2^4}{4} - \frac{0^4}{4} \right) = 2 \left( \frac{16}{4} \right) = 2 \times 4 = 8 \]

3. Add both parts:

\[ \frac{81}{2} + 8 = \frac{81}{2} + \frac{16}{2} = \frac{97}{2} \]

\[ \boxed{\frac{97}{2}} \]