What is the integral from 0 to 4 of (4x-3sqrt(x)) ?

The integral from 0 to 4 of (4x-3sqrt(x)) is 16

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integral from 0 to 4 of (4x-3sqrt(x))

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

16

Solution steps

\int_{0}^{4} 4 x - 3 x 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} 4 x d x - \int_{0}^{4} 3 x d x

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

\int_{0}^{4} 3 x d x = 16

= 32 - 16

Simplify

= 16

\int_{2}^{4} \frac{2 x}{x ^{2} - 1} d x

\int_{0}^{\infty} e

\int_{- 3}^{7} \frac{2 x + 6}{5} d x

\int_{- 5}^{5} - x^{2} + 25 d x

\int_{1}^{2} \frac{8}{x ^{3}} d x

What is the integral from 0 to 4 of (4x-3sqrt(x)) ?
The integral from 0 to 4 of (4x-3sqrt(x)) is 16