Solve ∫ (from 1 to 2) of v^5+5v^8/v^4dv

$$\int _ { 1 } ^ { 2 } \frac { v ^ { 5 } + 5 v ^ { 8 } } { v ^ { 4 } } d v$$

Evaluate

$$\frac{65}{2}=32.5$$

Solution Steps

Factor the expressions that are not already factored in \frac{v^{5}+5v^{8}}{v^{4}}.

$$\int _{1}^{2}\frac{\left(5v^{3}+1\right)v^{5}}{v^{4}}\mathrm{d}v$$

Cancel out v^{4} in both numerator and denominator.

$$\int _{1}^{2}v\left(5v^{3}+1\right)\mathrm{d}v$$

Expand the expression.

$$\int _{1}^{2}5v^{4}+v\mathrm{d}v$$

Evaluate the indefinite integral first.

$$\int 5v^{4}+v\mathrm{d}v$$

Integrate the sum term by term.

$$\int 5v^{4}\mathrm{d}v+\int v\mathrm{d}v$$

Factor out the constant in each of the terms.

$$5\int v^{4}\mathrm{d}v+\int v\mathrm{d}v$$

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

$$v^{5}+\int v\mathrm{d}v$$

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

$$v^{5}+\frac{v^{2}}{2}$$

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.

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

Simplify.

$$\frac{65}{2}$$