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Popular >

area y=x^2,y=x+2,0

Solution

area

y = x^{2}, y = x + 2, 0

Solution

No Solution

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

x^{2} = x + 2 : x = 2, x = - 1

x^{2} = 0 : x = 0

x + 2 = 0 : x = - 2

Curves not bounded

= No Solution

Popular Examples

area y= x/(x^2-4x+5),y= 3/(x^2-4x+5),x=0 area

y = \frac{x}{x ^{2} - 4 x + 5}, y = \frac{3}{x ^{2} - 4 x + 5}, x = 0

slope of y=(x^3+1)/(6x),x=1 slope

y = \frac{x ^{3} + 1}{6 x}, x = 1

y^'+4y=e^{-x},y(0)=10

y^{^{'}} + 4 y = e^{- x}, y (0) = 10

integral from 0 to 6 of 1/(sqrt(|x-3|))

\int_{0}^{6} \frac{1}{∣ x - 3 ∣} d x

derivative of x(x^2+3^3)

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

Frequently Asked Questions (FAQ)

What is area y=x^2,y=x+2,0 ?
The answer to area y=x^2,y=x+2,0 is No Solution