What is the integral from 0 to 4 of (6x+sqrt(x)) ?

The integral from 0 to 4 of (6x+sqrt(x)) is 160/3

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integral from 0 to 4 of (6x+sqrt(x))

\int_{0}^{4} (6 x + x) d x

\frac{160}{3}

Solution steps

\int_{0}^{4} 6 x + 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} 6 x d x + \int_{0}^{4} x d x

\int_{0}^{4} 6 x d x = 48

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

= 48 + \frac{16}{3}

Simplify

48 + \frac{16}{3} : \frac{160}{3}

= \frac{160}{3}

\int_{0}^{1} x^{2} - 2 x + 3 d x

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

\int_{1}^{2} \frac{10 - 5 x}{x ^{3}} d x

\int_{0}^{1} 3 y^{2} (e^{y^{3}} - 1) d y

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

What is the integral from 0 to 4 of (6x+sqrt(x)) ?
The integral from 0 to 4 of (6x+sqrt(x)) is 160/3