What is the integral from 0 to 60 of 10(1+x/(60))^2 ?

The integral from 0 to 60 of 10(1+x/(60))^2 is 1400

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integral from 0 to 60 of 10(1+x/(60))^2

\int_{0}^{60} 10 (1 + \frac{x}{60})^{2} d x

1400

Solution steps

\int_{0}^{60} 10 (1 + \frac{x}{60})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int_{0}^{60} (1 + \frac{x}{60})^{2} d x

Expand

(1 + \frac{x}{60})^{2} : 1 + \frac{x}{30} + \frac{x ^{2}}{3600}

= 10 \cdot \int_{0}^{60} 1 + \frac{x}{30} + \frac{x ^{2}}{3600} 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

= 10 (\int_{0}^{60} 1 d x + \int_{0}^{60} \frac{x}{30} d x + \int_{0}^{60} \frac{x ^{2}}{3600} d x)

\int_{0}^{60} 1 d x = 60

\int_{0}^{60} \frac{x}{30} d x = 60

\int_{0}^{60} \frac{x ^{2}}{3600} d x = 20

= 10 (60 + 60 + 20)

Simplify

= 1400

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What is the integral from 0 to 60 of 10(1+x/(60))^2 ?
The integral from 0 to 60 of 10(1+x/(60))^2 is 1400