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Popular >

area y=2x-2x^2,[0,1]

Solution

area

y = 2 x - 2 x^{2}, [0, 1]

Solution

\frac{1}{3}

+1

Solution steps

Apply the area formula:

\int_{0}^{1} 2 x - 2 x^{2} d x

Solve

\int_{0}^{1} 2 x - 2 x^{2} d x : \frac{1}{3}

The area is:

= \frac{1}{3}

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y = \frac{1}{x ^{2}}, y = 0, x = 1, x = 5

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x \to 0 lim (\frac{sin ( 2 x )}{x} + 4)

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

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x \to \infty lim (x^{2} + 3)

integral from 0 to 12 of sqrt(3x)

\int_{0}^{12} 3 x d x

Frequently Asked Questions (FAQ)

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