What is area y=12-x^2,y=-4x+0,[3,8] ?

The answer to area y=12-x^2,y=-4x+0,[3,8] is 137/3

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area y=12-x^2,y=-4x+0,[3,8]

area

y = 12 - x^{2}, y = - 4 x + 0, [3, 8]

\frac{137}{3}

Decimal Notation

45.66666 \dots

Solution steps

Apply the area formula:

\int_{3}^{8} 12 - x^{2} + 4 x d x

Solve

\int_{3}^{8} 12 - x^{2} + 4 x d x : \frac{137}{3}

The area is:

= \frac{137}{3}

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

\int sec (2 x) tan^{5} (2 x) d x

\int (e^{x y}) d y

(\frac{1}{tan ( x )})^{^{'}}

lo g_{5} (5 x^{3} + 2 x + 5)

What is area y=12-x^2,y=-4x+0,[3,8] ?
The answer to area y=12-x^2,y=-4x+0,[3,8] is 137/3