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

The integral from 1 to 4 of sqrt(x)(1+x) is 256/15

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

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

\frac{256}{15}

Decimal Notation

17.06666 \dots

Solution steps

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

Expand

x (1 + x) : x + x^{\frac{3}{2}}

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

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

\int_{1}^{4} x^{\frac{3}{2}} d x = \frac{62}{5}

= \frac{14}{3} + \frac{62}{5}

Simplify

\frac{14}{3} + \frac{62}{5} : \frac{256}{15}

= \frac{256}{15}

sin^{4} (x)

\int \frac{2 u ^{2} - 1}{u ^{3} - u} d u

\int_{100}^{120} 600 d x

What is the integral from 1 to 4 of sqrt(x)(1+x) ?
The integral from 1 to 4 of sqrt(x)(1+x) is 256/15