Solve ∫ (from - 1 to 1) of (5u^5+2u)du

$$\int _ { - 1 } ^ { 1 } ( 5 u ^ { 5 } + 2 u ) d u$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 5u^{5}+2u\mathrm{d}u$$

Integrate the sum term by term.

$$\int 5u^{5}\mathrm{d}u+\int 2u\mathrm{d}u$$

Factor out the constant in each of the terms.

$$5\int u^{5}\mathrm{d}u+2\int u\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u^{5}\mathrm{d}u with \frac{u^{6}}{6}. Multiply 5 times \frac{u^{6}}{6}.

$$\frac{5u^{6}}{6}+2\int u\mathrm{d}u$$

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

$$\frac{5u^{6}}{6}+u^{2}$$

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.

$$\frac{5}{6}\times 1^{6}+1^{2}-\left(\frac{5}{6}\left(-1\right)^{6}+\left(-1\right)^{2}\right)$$

Simplify.

$$0$$