Question

What is the definite integral of this |x^2 -3x-10| . limit is from -2 to +8?

Answer 1

Answer on Question #51701 – Math – Integral Calculus

What is the definite integral of this $|x^2 - 3x - 10|$, limit is from -2 to 8?

Solution

We must calculate the following integral:

$$\int_{-2}^{8} |x^2 - 3x - 10| \, dx.$$

To do this, we need to know how function $x^2 - 3x - 10$ behave on segment $[-2; 8]$:

$$x^2 - 3x - 10 = 0 \quad \Rightarrow \quad (x - 5)(x + 2) = 0.$$

We see that this function takes negative values on $[-2; 5]$ and takes positive values on $[5; 8]$. Now, using definition of absolute value $|A| = \begin{cases} A, & \text{if } A \geq 0 \\ -A, & \text{if } A < 0 \end{cases}$, we can rewrite the integral:

$$\begin{aligned} \int_{-2}^{8} |x^2 - 3x - 10| \, dx &= \int_{-2}^{5} (-x^2 + 3x + 10) \, dx + \int_{5}^{8} (x^2 - 3x - 10) \, dx = \left(-\frac{x^3}{3} + \frac{3}{2}x^2 + 10x\right) \Big|_{-2}^{5} + \left(\frac{x^3}{3} - \frac{3}{2}x^2 - 10x\right) \Big|_{5}^{8} \\ &= \left(-\frac{5^3}{3} + \frac{3}{2}5^2 + 10 \cdot 5\right) - \left(-\frac{(-2)^3}{3} + \frac{3}{2}(-2)^2 + 10 \cdot (-2)\right) \\ &\frac{3}{2} \\ \end{aligned}$$ $$\frac{3}{2}(-2)^2 + 10 \cdot (-2) + \left(\frac{8^3}{3} - \frac{3}{2}8^2 - 10 \cdot 8\right) - \left(\frac{5^3}{3} - \frac{3}{2}5^2 - 10 \cdot 5\right) = \frac{293}{3}.$$

Answer:

$$\int_{-2}^{8} |x^2 - 3x - 10| \, dx = \frac{293}{3}$$

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