Solve ∫ (from 0 to 10) of (n^2+n)dn

$$\int _ { 0 } ^ { 10 } ( n ^ { 2 } + n ) d n$$

Evaluate

$$\frac{1150}{3}\approx 383.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int n^{2}+n\mathrm{d}n$$

Integrate the sum term by term.

$$\int n^{2}\mathrm{d}n+\int n\mathrm{d}n$$

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

$$\frac{n^{3}}{3}+\int n\mathrm{d}n$$

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

$$\frac{n^{3}}{3}+\frac{n^{2}}{2}$$

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.

$$\frac{10^{3}}{3}+\frac{10^{2}}{2}-\left(\frac{0^{3}}{3}+\frac{0^{2}}{2}\right)$$

Simplify.

$$\frac{1150}{3}$$