What is the integral from 1 to 4 of (x-3/(sqrt(x))) ?

The integral from 1 to 4 of (x-3/(sqrt(x))) is 3/2

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integral from 1 to 4 of (x-3/(sqrt(x)))

\int_{1}^{4} (x - \frac{3}{x}) d x

\frac{3}{2}

Decimal

1.5

Solution steps

\int_{1}^{4} x - \frac{3}{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{3}{x} d x

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

\int_{1}^{4} \frac{3}{x} d x = 6

= \frac{15}{2} - 6

Simplify

\frac{15}{2} - 6 : \frac{3}{2}

= \frac{3}{2}

\int_{0}^{1} (2 - 2 x^{2}) d x

\int_{- 2}^{1} - y^{2} - y + 2 d y

\int_{0}^{2} (2 x^{3} - 2 x^{2} - 4 x - 1) d x

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

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

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