Solve ∫ (from 0 to 5) of (57n^5+97n^3+7n)dn

$$\int _ { 0 } ^ { 5 } ( 57 n ^ { 5 } + 97 n ^ { 3 } + 7 n ) d n$$

Evaluate

$$\frac{654725}{4}=163681.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 57n^{5}+97n^{3}+7n\mathrm{d}n$$

Integrate the sum term by term.

$$\int 57n^{5}\mathrm{d}n+\int 97n^{3}\mathrm{d}n+\int 7n\mathrm{d}n$$

Factor out the constant in each of the terms.

$$57\int n^{5}\mathrm{d}n+97\int n^{3}\mathrm{d}n+7\int n\mathrm{d}n$$

Since \int n^{k}\mathrm{d}n=\frac{n^{k+1}}{k+1} for k\neq -1, replace \int n^{5}\mathrm{d}n with \frac{n^{6}}{6}. Multiply 57 times \frac{n^{6}}{6}.

$$\frac{19n^{6}}{2}+97\int n^{3}\mathrm{d}n+7\int n\mathrm{d}n$$

Since \int n^{k}\mathrm{d}n=\frac{n^{k+1}}{k+1} for k\neq -1, replace \int n^{3}\mathrm{d}n with \frac{n^{4}}{4}. Multiply 97 times \frac{n^{4}}{4}.

$$\frac{19n^{6}}{2}+\frac{97n^{4}}{4}+7\int n\mathrm{d}n$$

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}. Multiply 7 times \frac{n^{2}}{2}.

$$\frac{19n^{6}}{2}+\frac{97n^{4}}{4}+\frac{7n^{2}}{2}$$

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.

$$\frac{19}{2}\times 5^{6}+\frac{97}{4}\times 5^{4}+\frac{7}{2}\times 5^{2}-\left(\frac{19}{2}\times 0^{6}+\frac{97}{4}\times 0^{4}+\frac{7}{2}\times 0^{2}\right)$$

Simplify.

$$\frac{654725}{4}$$