What is area x/(1+x^2),(x^2)/(1+x^3),[0,1] ?

The answer to area x/(1+x^2),(x^2)/(1+x^3),[0,1] is 1/6 ln(2)

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area x/(1+x^2),(x^2)/(1+x^3),[0,1]

area

\frac{x}{1 + x ^{2}}, \frac{x ^{2}}{1 + x ^{3}}, [0, 1]

\frac{1}{6} ln (2)

Decimal

0.11552 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{1} \frac{x}{1 + x ^{2}} - \frac{x ^{2}}{1 + x ^{3}} d x : \frac{1}{6} ln (2)

The area is:

= \frac{1}{6} ln (2)

\int \frac{8}{4 + 8 x} d x

\frac{180}{x ^{7}} - \frac{4}{x ^{3}}

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

f (x) = x^{2}, (0, - 9)

sin (θ) + cos (θ)

What is area x/(1+x^2),(x^2)/(1+x^3),[0,1] ?
The answer to area x/(1+x^2),(x^2)/(1+x^3),[0,1] is 1/6 ln(2)