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Popular >

area y^2=x+1,x=1

Solution

area

y^{2} = x + 1, x = 1

Solution

\frac{8 2}{3}

+1

Decimal Notation

3.77123 \dots

Solution steps

Isolate

y

for

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

Apply the area formula:

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

Solve

\int_{- 1}^{1} 2 x + 1 d x : \frac{8 2}{3}

The area is:

= \frac{8 2}{3}

Popular Examples

integral of sin^{10}(x)

\int sin^{10} (x) d x

tangent of f(x)= 4/(x^2),(1,4)tangent of

f (x) = \frac{4}{x ^{2}}, (1, 4)

derivative of f(x)=(3-e^{-4x})^4 derivative of

f (x) = (3 - e^{- 4 x})^{4}

derivative of sqrt(2+2cos(x))

\frac{d}{d x} (2 + 2 cos (x))

area 29-11x,1-x^2 area

29 - 11 x, 1 - x^{2}

Frequently Asked Questions (FAQ)

What is area y^2=x+1,x=1 ?
The answer to area y^2=x+1,x=1 is (8sqrt(2))/3