Solve ∫ (from -2 to 5) of (8x^3-2x) wrt x

$$\int_{ -2 }^{ 5 } (8 { x }^{ 3 } -2x) d x$$

Evaluate

$$1197$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8x^{3}-2x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int -2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x-2\int x\mathrm{d}x$$

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 8 times \frac{x^{4}}{4}.

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

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 -2 times \frac{x^{2}}{2}.

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

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.

$$2\times 5^{4}-5^{2}-\left(2\left(-2\right)^{4}-\left(-2\right)^{2}\right)$$

Simplify.

$$1197$$