Question

Will mark brainliest

Answer 1

The plot of $\sqrt{25-x^2}$ is that of the top half of a circle with radius 5. The interval [-5, 0] captures the left half of this semicircle.

Adding 5 to this shifts the plot up by 5 units. The area under this curve is then the combined area of a square with side length 5 and the area of a quarter circle with radius 5.

So we have

$\displaystyle \int_{-5}^0 5 + \sqrt{25-x^2} \, dx = 5^2 + \frac\pi4\cdot5^2 = \boxed{25 + \frac{25\pi}4}$