What is area y=x,g(y,x)=x^2,[0,1] ?

The answer to area y=x,g(y,x)=x^2,[0,1] is 1/6

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area y=x,g(y,x)=x^2,[0,1]

area

y = x, g (y, x) = x^{2}, [0, 1]

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{1} x - x^{2} d x : \frac{1}{6}

The area is:

= \frac{1}{6}

\frac{d}{d x} (cos (sin^{3} (5 x - 5)))

y^{^{'}} = \frac{( y ^{3} + 1 )}{( 3 x y ^{2} )}

x \to + 1 lim (\frac{1 - ∣ x ∣}{x - 1})

\int \frac{x ^{3}}{1 + x ^{4}} d x

y^{^{'}} - y + x^{2} = 0

What is area y=x,g(y,x)=x^2,[0,1] ?
The answer to area y=x,g(y,x)=x^2,[0,1] is 1/6