Solve ∫ (from 0 to 1) of 5u^5+3u^2+udu

$$\int _ { 0 } ^ { 1 } 5 u ^ { 5 } + 3 u ^ { 2 } + u d u$$

Evaluate

$$\frac{7}{3}\approx 2.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 5u^{5}+3u^{2}+u\mathrm{d}u$$

Integrate the sum term by term.

$$\int 5u^{5}\mathrm{d}u+\int 3u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

Factor out the constant in each of the terms.

$$5\int u^{5}\mathrm{d}u+3\int u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

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}.

$$\frac{5u^{6}}{6}+3\int u^{2}\mathrm{d}u+\int u\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u^{2}\mathrm{d}u with \frac{u^{3}}{3}. Multiply 3 times \frac{u^{3}}{3}.

$$\frac{5u^{6}}{6}+u^{3}+\int u\mathrm{d}u$$

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}.

$$\frac{5u^{6}}{6}+u^{3}+\frac{u^{2}}{2}$$

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.

$$\frac{5}{6}\times 1^{6}+1^{3}+\frac{1^{2}}{2}-\left(\frac{5}{6}\times 0^{6}+0^{3}+\frac{0^{2}}{2}\right)$$

Simplify.

$$\frac{7}{3}$$