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

The integral from 1 to 4 of (x+sqrt(x)) is 73/6

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

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

\frac{73}{6}

Decimal Notation

12.16666 \dots

Solution steps

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

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

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

= \frac{15}{2} + \frac{14}{3}

Simplify

\frac{15}{2} + \frac{14}{3} : \frac{73}{6}

= \frac{73}{6}

\int \frac{( x ^{4} + 81 )}{x ( x ^{2} + 9 ) ^{2}} d x

\frac{1}{s ^{2} - 16}

\int_{- \infty}^{0} e^{- x^{2}} d x

\int x \cdot ln^{4} (x) d x

x \to 2 - lim (- \frac{1}{x - 2})

What is the integral from 1 to 4 of (x+sqrt(x)) ?
The integral from 1 to 4 of (x+sqrt(x)) is 73/6