What is the integral from 1 to 2 of (v^4+5v^7)/(v^3) ?

The integral from 1 to 2 of (v^4+5v^7)/(v^3) is 65/2

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integral from 1 to 2 of (v^4+5v^7)/(v^3)

\int_{1}^{2} \frac{v ^{4} + 5 v ^{7}}{v ^{3}} d v

\frac{65}{2}

Decimal Notation

32.5

Solution steps

\int_{1}^{2} \frac{v ^{4} + 5 v ^{7}}{v ^{3}} d v

Expand

\frac{v ^{4} + 5 v ^{7}}{v ^{3}} : \frac{v ^{4}}{v ^{3}} + \frac{5 v ^{7}}{v ^{3}}

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 ^{4}}{v ^{3}} d v + \int_{1}^{2} \frac{5 v ^{7}}{v ^{3}} d v

\int_{1}^{2} \frac{v ^{4}}{v ^{3}} d v = \frac{3}{2}

\int_{1}^{2} \frac{5 v ^{7}}{v ^{3}} d v = 31

= \frac{3}{2} + 31

Simplify

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

= \frac{65}{2}

- t^{2} δ (t - 2)

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What is the integral from 1 to 2 of (v^4+5v^7)/(v^3) ?
The integral from 1 to 2 of (v^4+5v^7)/(v^3) is 65/2