What is area x=-2,x=3,y=2x^2+6,y=0 ?

The answer to area x=-2,x=3,y=2x^2+6,y=0 is 160/3

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area x=-2,x=3,y=2x^2+6,y=0

area

x = - 2, x = 3, y = 2 x^{2} + 6, y = 0

\frac{160}{3}

Decimal

53.33333 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{3} 2 x^{2} + 6 d x

Solve

\int_{- 2}^{3} 2 x^{2} + 6 d x : \frac{160}{3}

The area is:

= \frac{160}{3}

x \to 0 + lim (\frac{x ^{2} + 1}{x})

x \to 2 lim (\frac{1}{x ^{2} - 9})

y^{^{''}} + 15 y = 0

\frac{d}{d x} (f (x) (x)^{g (x)})

\frac{d}{d x} (\frac{( x + 1 ) ( 2 x - 5 )}{2 x + 3})

What is area x=-2,x=3,y=2x^2+6,y=0 ?
The answer to area x=-2,x=3,y=2x^2+6,y=0 is 160/3