What is area-x^3+x^2+16x,4x,0,4 ?

The answer to area-x^3+x^2+16x,4x,0,4 is 160/3

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area-x^3+x^2+16x,4x,0,4

area

- x^{3} + x^{2} + 16 x, 4 x, 0, 4

\frac{160}{3}

Decimal Notation

53.33333 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} - x^{3} + x^{2} + 12 x d x

Solve

\int_{0}^{4} - x^{3} + x^{2} + 12 x d x : \frac{160}{3}

The area is:

= \frac{160}{3}

f (x) = (4 x + 5)^{4}, at x = - 1

n = 1 \sum \infty \frac{n}{3 n + 1}

\frac{\partial}{\partial x} (\frac{1}{x} cos (\frac{y}{x}))

y^{^{'}} = (\frac{2 y + 3}{4 x + 5})^{2}

\int \frac{1}{( 25 x ^{2} + 4 ) ^{2}} d x

What is area-x^3+x^2+16x,4x,0,4 ?
The answer to area-x^3+x^2+16x,4x,0,4 is 160/3