What is the integral from 0 to 2 of 0.09375(4-x^2) ?

The integral from 0 to 2 of 0.09375(4-x^2) is 0.5

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integral from 0 to 2 of 0.09375(4-x^2)

\int_{0}^{2} 0.09375 (4 - x^{2}) d x

0.5

Solution steps

\int_{0}^{2} 0.09375 (4 - x^{2}) d x

Take the constant out:

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

= 0.09375 \cdot \int_{0}^{2} 4 - x^{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

= 0.09375 (\int_{0}^{2} 4 d x - \int_{0}^{2} x^{2} d x)

\int_{0}^{2} 4 d x = 8

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

= 0.09375 (8 - \frac{8}{3})

Simplify

0.09375 (8 - \frac{8}{3}) : 0.5

= 0.5

\int sec (6 w) tan (6 w) d w

\int 2 x - x d x

t an g e n t y = x + 5, at x = 11

\int_{- 1}^{4} x - 2 x^{2} d x

f (x) = 2^{s i n (x)}

What is the integral from 0 to 2 of 0.09375(4-x^2) ?
The integral from 0 to 2 of 0.09375(4-x^2) is 0.5