What is area y=x^2-1,y=-x+2,x=0,x=1 ?

The answer to area y=x^2-1,y=-x+2,x=0,x=1 is 13/6

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area y=x^2-1,y=-x+2,x=0,x=1

area

y = x^{2} - 1, y = - x + 2, x = 0, x = 1

\frac{13}{6}

Decimal Notation

2.16666 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{1} x^{2} - 1 - (- x + 2) d x : \frac{13}{6}

The area is:

= \frac{13}{6}

\int \frac{3}{x ^{4}} + 2 - \frac{3}{x ^{2}} d x

\frac{d}{d x} (9 x sin (x^{2}))

y^{^{''}} + 9 y^{^{'}} = 0

y^{^{'}} = 3 x^{2} y - 8 e^{(x^{3} - 4 x)}

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x), y (0) = 1

What is area y=x^2-1,y=-x+2,x=0,x=1 ?
The answer to area y=x^2-1,y=-x+2,x=0,x=1 is 13/6