What is area y=x^3,y=x^2-2x,[0,1] ?

The answer to area y=x^3,y=x^2-2x,[0,1] is 11/12

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area y=x^3,y=x^2-2x,[0,1]

area

y = x^{3}, y = x^{2} - 2 x, [0, 1]

\frac{11}{12}

Decimal Notation

0.91666 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} x^{3} - (x^{2} - 2 x) d x

Solve

\int_{0}^{1} x^{3} - (x^{2} - 2 x) d x : \frac{11}{12}

The area is:

= \frac{11}{12}

\int_{0}^{1} \frac{y}{e ^{8 y}} d y

\int \frac{ln ( 7 x )}{x} d x

\frac{3 s ^{2}}{( s + 2 ) ( s + 5 )}

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

f (x) = (4 x + 3)^{2}

What is area y=x^3,y=x^2-2x,[0,1] ?
The answer to area y=x^3,y=x^2-2x,[0,1] is 11/12