Question

Evaluate the integral: \[ \int_0^2 |x^2 + x - 2| \, dx \]

• A. \( \frac{11}{3} \)
• B. \( \frac{-11}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{-1}{3} \)

Hint:

When working with absolute values in integrals, always split the integral at the points where the function inside the absolute value changes sign.

Answer 1

The Correct Option is A

We are asked to evaluate the integral of the absolute value function. To handle the absolute value, we first find the points where the expression inside the absolute value changes sign. This happens when the quadratic expression \( x^2 + x - 2 \) equals zero.
Step 1: Find the roots of \( x^2 + x - 2 = 0 \).
We can solve for \( x \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 1 \), and \( c = -2 \). Thus, \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}. \] So the roots are: \[ x = \frac{-1 + 3}{2} = 1 \quad \text{and} \quad x = \frac{-1 - 3}{2} = -2. \]
Step 2: Determine the sign of \( x^2 + x - 2 \) on different intervals.
We now examine the sign of \( x^2 + x - 2 \) in the intervals \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \): - For \( x<-2 \), the quadratic expression is positive because both factors \( (x - 1) \) and \( (x + 2) \) are negative, making their product positive. - For \( -2<x<1 \), the quadratic expression is negative because \( (x - 1) \) is negative, and \( (x + 2) \) is positive. - For \( x>1 \), the quadratic expression is positive because both factors \( (x - 1) \) and \( (x + 2) \) are positive. Thus, we have: \[ |x^2 + x - 2| = \begin{cases} x^2 + x - 2, & \text{for} \, x \in (-\infty, -2) \cup (1, \infty) -(x^2 + x - 2), & \text{for} \, x \in (-2, 1) \end{cases} \]
Step 3: Split the integral at \( x = 1 \) and \( x = -2 \).
We now split the integral as follows: \[ \int_0^2 |x^2 + x - 2| \, dx = \int_0^1 |x^2 + x - 2| \, dx + \int_1^2 |x^2 + x - 2| \, dx. \] For \( 0<x<1 \), \( x^2 + x - 2 \) is negative, so: \[ |x^2 + x - 2| = -(x^2 + x - 2). \] Thus, the integral from 0 to 1 is: \[ \int_0^1 -(x^2 + x - 2) \, dx = -\int_0^1 (x^2 + x - 2) \, dx. \] For \( 1<x<2 \), \( x^2 + x - 2 \) is positive, so: \[ |x^2 + x - 2| = x^2 + x - 2. \] Thus, the integral from 1 to 2 is: \[ \int_1^2 (x^2 + x - 2) \, dx. \]
Step 4: Evaluate the integrals.
We now evaluate both integrals: \[ \int_0^1 (x^2 + x - 2) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_0^1 = \left( \frac{1}{3} + \frac{1}{2} - 2 \right) - 0 = \frac{1}{3} + \frac{1}{2} - 2 = -\frac{3}{6} = -\frac{1}{3}. \] Thus, the first integral becomes: \[ -\left( -\frac{1}{3} \right) = \frac{1}{3}. \] Next, we evaluate the second integral: \[ \int_1^2 (x^2 + x - 2) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_1^2 = \left( \frac{8}{3} + \frac{4}{2} - 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) \] \[ = \left( \frac{8}{3} + 2 - 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = \frac{8}{3} - 2 - 4 = \frac{8}{3} - \frac{18}{3} = -\frac{10}{3}. \] The second integral gives: \[ -\frac{10}{3}. \]
Step 5: Add the results of the two integrals.
Thus, the total value of the integral is: \[ \frac{1}{3} + \left( -\frac{10}{3} \right) = \frac{11}{3}. \]