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Popular >

area y=6,y=x^{-2}+x^2

Solution

area

y = 6, y = x^{- 2} + x^{2}

Solution

\frac{16}{3}

+1

Solution steps

Apply the area formula:

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

Solve

\int_{2 - 1}^{2 + 1} 6 - (\frac{1}{x ^{2}} + x^{2}) d x : \frac{16}{3}

The area is:

= \frac{16}{3}

Popular Examples

integral of e^{(3-x)}

\int e^{(3 - x)} d x

x^{^{''}} + x = 0

derivative of 2x^2+1 derivative of

2 x^{2} + 1

integral of 1/2 (e^x+e^{-x})

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

(dy)/(dx)(x)=y+(x^2-3)^2

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

Frequently Asked Questions (FAQ)

What is area y=6,y=x^{-2}+x^2 ?
The answer to area y=6,y=x^{-2}+x^2 is 16/3