What is the integral from-1 to 1 of 0.09375(4-x^2) ?

The integral from-1 to 1 of 0.09375(4-x^2) is 0.6875

Popular >

integral from-1 to 1 of 0.09375(4-x^2)

\int_{- 1}^{1} 0.09375 (4 - x^{2}) d x

0.6875

Solution steps

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

\int_{- 1}^{1} 4 d x = 8

\int_{- 1}^{1} x^{2} d x = \frac{2}{3}

= 0.09375 (8 - \frac{2}{3})

Simplify

0.09375 (8 - \frac{2}{3}) : 0.6875

= 0.6875

\int e^{- 17 x} d x

\int \frac{x}{16 - x ^{8}} d x

\frac{d y}{d x} = \frac{2 x + y}{3 + 3 y ^{2} - x}, y (0) = 0

\frac{d}{d x} (ln (sin (x)))

(3, 5), (- 2, - 20)

What is the integral from-1 to 1 of 0.09375(4-x^2) ?
The integral from-1 to 1 of 0.09375(4-x^2) is 0.6875