What is the integral from 1 to 4 of (2x^2+sqrt(x)) ?

The integral from 1 to 4 of (2x^2+sqrt(x)) is 140/3

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integral from 1 to 4 of (2x^2+sqrt(x))

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

\frac{140}{3}

Decimal

46.66666 \dots

Solution steps

\int_{1}^{4} 2 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_{1}^{4} 2 x^{2} d x + \int_{1}^{4} x d x

\int_{1}^{4} 2 x^{2} d x = 42

\int_{1}^{4} x d x = \frac{14}{3}

= 42 + \frac{14}{3}

Simplify

42 + \frac{14}{3} : \frac{140}{3}

= \frac{140}{3}

\int_{0}^{\frac{π}{4}} arctan (x) d x

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

What is the integral from 1 to 4 of (2x^2+sqrt(x)) ?
The integral from 1 to 4 of (2x^2+sqrt(x)) is 140/3