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

The integral from 1 to 4 of (x^2-sqrt(x))/x is 11/2

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

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

\frac{11}{2}

Decimal

5.5

Solution steps

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

Expand

\frac{x ^{2} - x}{x} : x - \frac{1}{x}

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

\int_{1}^{4} x d x = \frac{15}{2}

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

= \frac{15}{2} - 2

Simplify

\frac{15}{2} - 2 : \frac{11}{2}

= \frac{11}{2}

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What is the integral from 1 to 4 of (x^2-sqrt(x))/x ?
The integral from 1 to 4 of (x^2-sqrt(x))/x is 11/2