Solve ∫ (from 0 to 1) of (1-8v^3+16v^7)dv

$$\int _ { 0 } ^ { 1 } ( 1 - 8 v ^ { 3 } + 16 v ^ { 7 } ) d v$$

Evaluate

$$1$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1-8v^{3}+16v^{7}\mathrm{d}v$$

Integrate the sum term by term.

$$\int 1\mathrm{d}v+\int -8v^{3}\mathrm{d}v+\int 16v^{7}\mathrm{d}v$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}v=av.

$$v-8\int v^{3}\mathrm{d}v+16\int v^{7}\mathrm{d}v$$

Since \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} for k\neq -1, replace \int v^{3}\mathrm{d}v with \frac{v^{4}}{4}. Multiply -8 times \frac{v^{4}}{4}.

$$v-2v^{4}+16\int v^{7}\mathrm{d}v$$

Since \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} for k\neq -1, replace \int v^{7}\mathrm{d}v with \frac{v^{8}}{8}. Multiply 16 times \frac{v^{8}}{8}.

$$v-2v^{4}+2v^{8}$$

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.

$$1-2\times 1^{4}+2\times 1^{8}-\left(0-2\times 0^{4}+2\times 0^{8}\right)$$

Simplify.

$$1$$