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

The integral from-2 to 2 of 1/2 (4-x^2) is 16/3

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

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2}^{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

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

\int_{- 2}^{2} 4 d x = 16

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

= \frac{1}{2} (16 - \frac{16}{3})

Simplify

\frac{1}{2} (16 - \frac{16}{3}) : \frac{16}{3}

= \frac{16}{3}

\frac{3}{5 x + 2}, at x = - 1

\int_{0}^{1} \frac{6 - x}{1 + x ^{2}} d x

x \to 2 lim (x + 2)

\int \frac{1}{( 361 + x ^{2} ) ^{\frac{3}{2}}} d x

What is the integral from-2 to 2 of 1/2 (4-x^2) ?
The integral from-2 to 2 of 1/2 (4-x^2) is 16/3