Solve ∫ (from -1 to 1) of 8-8y^4 wrt y

$$\int_{ -1 }^{ 1 } 8-8 { y }^{ 4 } d y$$

Evaluate

$$\frac{64}{5}=12.8$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8-8y^{4}\mathrm{d}y$$

Integrate the sum term by term.

$$\int 8\mathrm{d}y+\int -8y^{4}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int 8\mathrm{d}y-8\int y^{4}\mathrm{d}y$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$8y-8\int y^{4}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{4}\mathrm{d}y with \frac{y^{5}}{5}. Multiply -8 times \frac{y^{5}}{5}.

$$8y-\frac{8y^{5}}{5}$$

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.

$$8\times 1-\frac{8}{5}\times 1^{5}-\left(8\left(-1\right)-\frac{8}{5}\left(-1\right)^{5}\right)$$

Simplify.

$$\frac{64}{5}$$