What is the integral from 0 to 4 of (3x^2+x-8) ?

The integral from 0 to 4 of (3x^2+x-8) is 40

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integral from 0 to 4 of (3x^2+x-8)

\int_{0}^{4} (3 x^{2} + x - 8) d x

40

Solution steps

\int_{0}^{4} 3 x^{2} + x - 8 d x

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} 3 x^{2} d x + \int_{0}^{4} x d x - \int_{0}^{4} 8 d x

\int_{0}^{4} 3 x^{2} d x = 64

\int_{0}^{4} x d x = 8

\int_{0}^{4} 8 d x = 32

= 64 + 8 - 32

Simplify

= 40

\int_{1}^{4} x^{3} d x

\int_{3}^{9} 1 + 3 x d x

\int_{0}^{5} \frac{1}{1 + x ^{2} + y ^{2}} d z

\int_{y}^{2 y} 8 x + e^{y} d x

What is the integral from 0 to 4 of (3x^2+x-8) ?
The integral from 0 to 4 of (3x^2+x-8) is 40