What is area y=(x^2)/2-3,y=-(x^2)/2+2x ?

The answer to area y=(x^2)/2-3,y=-(x^2)/2+2x is 32/3

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area y=(x^2)/2-3,y=-(x^2)/2+2x

area

y = \frac{x ^{2}}{2} - 3, y = - \frac{x ^{2}}{2} + 2 x

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 1}^{3} \frac{x ^{2}}{2} - 3 - (- \frac{x ^{2}}{2} + 2 x) d x : \frac{32}{3}

The area is:

= \frac{32}{3}

\int_{1}^{9} \frac{1}{x - 7} d x

\frac{d}{d x} (- x e^{(\frac{x + 1}{x - 1})})

\int \frac{x ^{2}}{x ^{2} - 4 x + 5} d x

\frac{d}{d x} (x + 5 e^{x})

What is area y=(x^2)/2-3,y=-(x^2)/2+2x ?
The answer to area y=(x^2)/2-3,y=-(x^2)/2+2x is 32/3