What is the integral from 1 to 2 of (v^3+5v^6)/(v^2) ?

The integral from 1 to 2 of (v^3+5v^6)/(v^2) is 65/2

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integral from 1 to 2 of (v^3+5v^6)/(v^2)

\int_{1}^{2} \frac{v ^{3} + 5 v ^{6}}{v ^{2}} d v

\frac{65}{2}

Solution steps

\int_{1}^{2} \frac{v ^{3} + 5 v ^{6}}{v ^{2}} d v

Expand

\frac{v ^{3} + 5 v ^{6}}{v ^{2}} : \frac{v ^{3}}{v ^{2}} + \frac{5 v ^{6}}{v ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{v ^{3}}{v ^{2}} d v + \int_{1}^{2} \frac{5 v ^{6}}{v ^{2}} d v

\int_{1}^{2} \frac{v ^{3}}{v ^{2}} d v = \frac{3}{2}

\int_{1}^{2} \frac{5 v ^{6}}{v ^{2}} d v = 31

= \frac{3}{2} + 31

Simplify

\frac{3}{2} + 31 : \frac{65}{2}

= \frac{65}{2}

\int_{2}^{5} ln (x^{2} - 2) d x

\int_{0}^{l n (x)} (x e^{- y}) d y

\int_{- \infty}^{2} e^{- 5 x} d x

\int_{- 2}^{1} \frac{1}{x ^{3}} d x

\int_{0}^{0} e^{- x^{2}} d x

What is the integral from 1 to 2 of (v^3+5v^6)/(v^2) ?
The integral from 1 to 2 of (v^3+5v^6)/(v^2) is 65/2