Solve ∫ (from - 8 to 1) of (64-x^2) wrt x

$$\int _ { - 8 } ^ { 1 } ( 64 - x ^ { 2 } ) d x$$

Evaluate

$$405$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 64-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 64\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$64x-\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -1 times \frac{x^{3}}{3}.

$$64x-\frac{x^{3}}{3}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$64\times 1-\frac{1^{3}}{3}-\left(64\left(-8\right)-\frac{\left(-8\right)^{3}}{3}\right)$$

Simplify.

$$405$$