What is the integral from 1 to 4 of sqrt(9x)-x ?

The integral from 1 to 4 of sqrt(9x)-x is 13/2

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integral from 1 to 4 of sqrt(9x)-x

\int_{1}^{4} 9 x - x d x

\frac{13}{2}

Decimal Notation

6.5

Solution steps

\int_{1}^{4} 9 x - x d x

9 x = 3 x

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

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

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

= 14 - \frac{15}{2}

Simplify

14 - \frac{15}{2} : \frac{13}{2}

= \frac{13}{2}

\int_{x}^{π} cos (t^{3}) d t

\int_{0}^{\frac{π}{4}} x d x

\int_{1 - x}^{1} e^{x} d x

\int_{0}^{5} (x - 1)^{2} d x

\int_{1}^{5} \frac{1}{x ^{2} - 1} d x

What is the integral from 1 to 4 of sqrt(9x)-x ?
The integral from 1 to 4 of sqrt(9x)-x is 13/2