Answer
$- \frac{25\pi}{2}$
Work Step by Step
$\int_{-5}^{5} (x-\sqrt{25-x^2})~dx = \int_{-5}^{5} x~dx-\int_{-5}^{5} \sqrt{25-x^2}~dx$
On the interval $-5 \leq x \leq 5$, the graph of $f(x) = x$ forms two triangles. One triangle is below the x-axis and one triangle is above the x-axis. We can find the area of these two triangles:
below the x-axis:
$A = \frac{1}{2}(5)(5) = \frac{25}{2}$
above the x-axis:
$A = \frac{1}{2}(5)(5) = \frac{25}{2}$
On the interval $-5 \leq x \leq 5$, the graph of $\sqrt{25-x^2}$ forms a semi-circle above the x-axis. We can find the area:
$A = \frac{\pi~r^2}{2} = \frac{25\pi}{2}$
Therefore:
$\int_{-5}^{5} (x-\sqrt{25-x^2})~dx = (\frac{25}{2}-\frac{25}{2})- \frac{25\pi}{2}= - \frac{25\pi}{2}$
