What is the integral from 1 to 3 of (x+1)^4-(x+1)^3 ?

The integral from 1 to 3 of (x+1)^4-(x+1)^3 is 692/5

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integral from 1 to 3 of (x+1)^4-(x+1)^3

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

\frac{692}{5}

Decimal

138.4

Solution steps

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

\int_{1}^{3} (x + 1)^{4} d x = \frac{992}{5}

\int_{1}^{3} (x + 1)^{3} d x = 60

= \frac{992}{5} - 60

Simplify

\frac{992}{5} - 60 : \frac{692}{5}

= \frac{692}{5}

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What is the integral from 1 to 3 of (x+1)^4-(x+1)^3 ?
The integral from 1 to 3 of (x+1)^4-(x+1)^3 is 692/5