What is the integral from 0 to 1 of (8x+sqrt(x)) ?

The integral from 0 to 1 of (8x+sqrt(x)) is 14/3

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integral from 0 to 1 of (8x+sqrt(x))

\int_{0}^{1} (8 x + x) d x

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{0}^{1} 8 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}^{1} 8 x d x + \int_{0}^{1} x d x

\int_{0}^{1} 8 x d x = 4

\int_{0}^{1} x d x = \frac{2}{3}

= 4 + \frac{2}{3}

Simplify

4 + \frac{2}{3} : \frac{14}{3}

= \frac{14}{3}

\int_{3}^{x^{2}} ln (t) d t

\int_{0}^{4} \frac{1}{2} d x

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

\int_{6.3}^{6.6} 12 x^{2} - 6 x + \frac{4}{x} d x

\int_{0}^{2} 12 x^{3} d x

What is the integral from 0 to 1 of (8x+sqrt(x)) ?
The integral from 0 to 1 of (8x+sqrt(x)) is 14/3