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Popular >

area y=2x^2,y=8x-2x^2

Solution

area

y = 2 x^{2}, y = 8 x - 2 x^{2}

Solution

\frac{16}{3}

+1

Decimal Notation

5.33333 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} 2 x^{2} - (8 x - 2 x^{2}) d x

Solve

\int_{0}^{2} 2 x^{2} - (8 x - 2 x^{2}) d x : \frac{16}{3}

The area is:

= \frac{16}{3}

Popular Examples

(\partial)/(\partial y)(x^2+y^2-13)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 13)

tangent of f(x)=-x^2-7x+44,\at x=2 tangent of

f (x) = - x^{2} - 7 x + 44, at x = 2

integral of-e^{3x}

\int - e^{3 x} d x

derivative of (2(-x^2+1))/((x^2+1)^2)derivative of

\frac{2 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

integral of y/(xz)

\int \frac{y}{x z} d x

Frequently Asked Questions (FAQ)

What is area y=2x^2,y=8x-2x^2 ?
The answer to area y=2x^2,y=8x-2x^2 is 16/3