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

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_{0}^{1} t d t

\int_{0}^{2} \frac{x ^{2} - 6 + x}{x + 3} d x

\int_{1}^{4} (x^{2} - 4 x + 4) d x

\int_{x}^{x^{2}} (e^{x^{4}}) d x

\int_{0}^{1} \frac{2 x + 1}{1 + x ^{2}} 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