What is the integral from 0 to 8 of (x+1)^2 ?

The integral from 0 to 8 of (x+1)^2 is 728/3

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integral from 0 to 8 of (x+1)^2

\int_{0}^{8} (x + 1)^{2} d x

\frac{728}{3}

Decimal

242.66666 \dots

Solution steps

\int_{0}^{8} (x + 1)^{2} d x

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

= \int_{0}^{8} x^{2} + 2 x + 1 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_{0}^{8} x^{2} d x + \int_{0}^{8} 2 x d x + \int_{0}^{8} 1 d x

\int_{0}^{8} x^{2} d x = \frac{512}{3}

\int_{0}^{8} 2 x d x = 64

\int_{0}^{8} 1 d x = 8

= \frac{512}{3} + 64 + 8

Simplify

\frac{512}{3} + 64 + 8 : \frac{728}{3}

= \frac{728}{3}

y = arcsin (\frac{1}{x ^{5}})

x \to 0 lim (\frac{ln ( 1 - 6 x )}{x})

x \to - 8 lim (\frac{∣ x + 8 ∣}{x + 8})

- \frac{4 x}{y}

f (x) = x cos (x)

What is the integral from 0 to 8 of (x+1)^2 ?
The integral from 0 to 8 of (x+1)^2 is 728/3