What is the integral from 1 to 4 of (x^{1/2}-x^2) ?

The integral from 1 to 4 of (x^{1/2}-x^2) is -49/3

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integral from 1 to 4 of (x^{1/2}-x^2)

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

- \frac{49}{3}

Decimal Notation

- 16.33333 \dots

Solution steps

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

\int_{1}^{4} x^{\frac{1}{2}} d x = \frac{14}{3}

\int_{1}^{4} x^{2} d x = 21

= \frac{14}{3} - 21

Simplify

\frac{14}{3} - 21 : - \frac{49}{3}

= - \frac{49}{3}

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\int_{- 8}^{7} (1 - x) d x

\int_{- 1}^{0} \frac{1}{1 + e ^{t}} d t

What is the integral from 1 to 4 of (x^{1/2}-x^2) ?
The integral from 1 to 4 of (x^{1/2}-x^2) is -49/3