Question

$\int_{-2}^{4} |2-x^2| dx =$

• A. $\frac{8\sqrt{2}-3}{3}$
• B. $\frac{16\sqrt{2}}{3}+12$
• C. $\frac{16\sqrt{2}-3}{3}$
• D. $\frac{8\sqrt{2}+12}{3}$

Hint:

When integrating an absolute value function $|f(x)|$, first find the roots of $f(x)=0$. These roots are the points where you need to split the interval of integration. Then, for each sub-interval, determine the sign of $f(x)$ to remove the absolute value bars correctly.

Answer 1

The Correct Option is B

To evaluate an integral with an absolute value, we must split the integral based on where the expression inside the absolute value changes sign.
The expression is $|2-x^2|$. It is zero when $x^2=2$, i.e., at $x=\pm\sqrt{2}$.
The parabola $y=2-x^2$ is positive between $-\sqrt{2}$ and $\sqrt{2}$, and negative otherwise.
The interval of integration is $[-2, 4]$. The points $\pm\sqrt{2}$ are within this interval. ($\sqrt{2} \approx 1.414$)
So, we split the integral into three parts:
$I = \int_{-2}^{-\sqrt{2}} -(2-x^2) dx + \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2) dx + \int_{\sqrt{2}}^{4} -(2-x^2) dx$.
$I = \int_{-2}^{-\sqrt{2}} (x^2-2) dx + \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2) dx + \int_{\sqrt{2}}^{4} (x^2-2) dx$.
Evaluate the first integral:
$[\frac{x^3}{3}-2x]_{-2}^{-\sqrt{2}} = (\frac{-2\sqrt{2}}{3}+2\sqrt{2}) - (\frac{-8}{3}+4) = \frac{4\sqrt{2}}{3} - \frac{4}{3}$.
Evaluate the second integral (note it's an even function over a symmetric interval):
$2\int_{0}^{\sqrt{2}} (2-x^2) dx = 2[2x-\frac{x^3}{3}]_{0}^{\sqrt{2}} = 2(2\sqrt{2}-\frac{2\sqrt{2}}{3}) = 2(\frac{4\sqrt{2}}{3}) = \frac{8\sqrt{2}}{3}$.
Evaluate the third integral:
$[\frac{x^3}{3}-2x]_{\sqrt{2}}^{4} = (\frac{64}{3}-8) - (\frac{2\sqrt{2}}{3}-2\sqrt{2}) = (\frac{64-24}{3}) - (\frac{2\sqrt{2}-6\sqrt{2}}{3}) = \frac{40}{3} - (-\frac{4\sqrt{2}}{3}) = \frac{40}{3} + \frac{4\sqrt{2}}{3}$.
Sum the three parts:
$I = \left(\frac{4\sqrt{2}}{3} - \frac{4}{3}\right) + \frac{8\sqrt{2}}{3} + \left(\frac{40}{3} + \frac{4\sqrt{2}}{3}\right)$.
$I = \frac{4\sqrt{2} - 4 + 8\sqrt{2} + 40 + 4\sqrt{2}}{3} = \frac{(4+8+4)\sqrt{2} + (40-4)}{3} = \frac{16\sqrt{2} + 36}{3}$.
$I = \frac{16\sqrt{2}}{3} + \frac{36}{3} = \frac{16\sqrt{2}}{3} + 12$.