Solve ∫ (from - 2 to 7) of (8x-x^3) wrt x

$$\int _ { - 2 } ^ { 7 } ( 8 x - x ^ { 3 } ) d x$$

Evaluate

$$-\frac{1665}{4}=-416.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8x-x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x\mathrm{d}x+\int -x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x\mathrm{d}x-\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 8 times \frac{x^{2}}{2}.

$$4x^{2}-\int x^{3}\mathrm{d}x$$

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}. Multiply -1 times \frac{x^{4}}{4}.

$$4x^{2}-\frac{x^{4}}{4}$$

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.

$$4\times 7^{2}-\frac{7^{4}}{4}-\left(4\left(-2\right)^{2}-\frac{\left(-2\right)^{4}}{4}\right)$$

Simplify.

$$-\frac{1665}{4}$$