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

The integral from 0 to 4 of (4x^2+sqrt(x)) is 272/3

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integral from 0 to 4 of (4x^2+sqrt(x))

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

\frac{272}{3}

Decimal Notation

90.66666 \dots

Solution steps

\int_{0}^{4} 4 x^{2} + 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^{2} d x + \int_{0}^{4} x d x

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

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

= \frac{256}{3} + \frac{16}{3}

Simplify

\frac{256}{3} + \frac{16}{3} : \frac{272}{3}

= \frac{272}{3}

\int_{1}^{5} \frac{8}{x} d x

\int_{- 5}^{4} x^{2} + 5 - x d x

\int_{1}^{x} \frac{1}{1 + t ^{2}} d t

\int_{0}^{\frac{π}{4}} sin^{2} (\frac{x}{2}) d x

\int_{0}^{\frac{π}{2}} 8 + 4 cos (x) d x

What is the integral from 0 to 4 of (4x^2+sqrt(x)) ?
The integral from 0 to 4 of (4x^2+sqrt(x)) is 272/3