What is the integral from 0 to 5 of (140x^2+9800) ?

The integral from 0 to 5 of (140x^2+9800) is 164500/3

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integral from 0 to 5 of (140x^2+9800)

\int_{0}^{5} (140 x^{2} + 9800) d x

\frac{164500}{3}

Decimal

54833.33333 \dots

Solution steps

\int_{0}^{5} 140 x^{2} + 9800 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}^{5} 140 x^{2} d x + \int_{0}^{5} 9800 d x

\int_{0}^{5} 140 x^{2} d x = \frac{17500}{3}

\int_{0}^{5} 9800 d x = 49000

= \frac{17500}{3} + 49000

Simplify

\frac{17500}{3} + 49000 : \frac{164500}{3}

= \frac{164500}{3}

What is the integral from 0 to 5 of (140x^2+9800) ?
The integral from 0 to 5 of (140x^2+9800) is 164500/3