Solve ∫ (from 0 to 4) of (x^3-16) wrt x

$$\int _ { 0 } ^ { 4 } ( x ^ { 3 } - 16 ) d x$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-16\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -16\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+\int -16\mathrm{d}x$$

Find the integral of -16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-16x$$

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.

$$\frac{4^{4}}{4}-16\times 4-\left(\frac{0^{4}}{4}-16\times 0\right)$$

Simplify.

$$0$$