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Popular >

area y=-1/2 x^2+2,y=0

Solution

area

y = - \frac{1}{2} x^{2} + 2, y = 0

Solution

\frac{16}{3}

+1

Decimal Notation

5.33333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 2}^{2} - \frac{1}{2} x^{2} + 2 d x : \frac{16}{3}

The area is:

= \frac{16}{3}

Popular Examples

integral of \sqrt[9]{e^x}

(y^2+1)dx=ysec^2(x)dy

(y^{2} + 1) d x = y sec^{2} (x) d y

limit as x approaches 0 of x^{-2}

x \to 0 lim (x^{- 2})

(2x)dx+(2y-4x^2y^{-1})dy=0

(2 x) d x + (2 y - 4 x^{2} y^{- 1}) d y = 0

derivative of ln|x^2-1|

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

Frequently Asked Questions (FAQ)

What is area y=-1/2 x^2+2,y=0 ?
The answer to area y=-1/2 x^2+2,y=0 is 16/3