What is the integral from 0 to 4 of (2v+5)(3v-1) ?

The integral from 0 to 4 of (2v+5)(3v-1) is 212

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integral from 0 to 4 of (2v+5)(3v-1)

\int_{0}^{4} (2 v + 5) (3 v - 1) d v

212

Solution steps

\int_{0}^{4} (2 v + 5) (3 v - 1) d v

Expand

(2 v + 5) (3 v - 1) : 6 v^{2} + 13 v - 5

= \int_{0}^{4} 6 v^{2} + 13 v - 5 d v

Apply the Sum Rule:

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

= \int_{0}^{4} 6 v^{2} d v + \int_{0}^{4} 13 v d v - \int_{0}^{4} 5 d v

\int_{0}^{4} 6 v^{2} d v = 128

\int_{0}^{4} 13 v d v = 104

\int_{0}^{4} 5 d v = 20

= 128 + 104 - 20

Simplify

= 212

\int \frac{x ^{2}}{6 x ^{2} - 49} d x

f (x) = x^{2} - 6, (- 4, 10)

\int (x^{- 3}) d x

\frac{d y}{d t} = t^{2} e^{- 4 t} - 4 y

7 x y^{3} + 5 x y = 12, at x = 1

What is the integral from 0 to 4 of (2v+5)(3v-1) ?
The integral from 0 to 4 of (2v+5)(3v-1) is 212