What is the integral from 0 to 1 of (4x+sqrt(x)) ?

The integral from 0 to 1 of (4x+sqrt(x)) is 8/3

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integral from 0 to 1 of (4x+sqrt(x))

\int_{0}^{1} (4 x + x) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{0}^{1} 4 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} 4 x d x + \int_{0}^{1} x d x

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

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

= 2 + \frac{2}{3}

Simplify

2 + \frac{2}{3} : \frac{8}{3}

= \frac{8}{3}

What is the integral from 0 to 1 of (4x+sqrt(x)) ?
The integral from 0 to 1 of (4x+sqrt(x)) is 8/3