What is area f(x)=8x+2x^2-x^3,g(x)=0 ?

The answer to area f(x)=8x+2x^2-x^3,g(x)=0 is 148/3

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area f(x)=8x+2x^2-x^3,g(x)=0

area

f (x) = 8 x + 2 x^{2} - x^{3}, g (x) = 0

\frac{148}{3}

Decimal Notation

49.33333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 2}^{4} - x^{3} + 2 x^{2} + 8 x d x : \frac{148}{3}

The area is:

= \frac{148}{3}

\int \frac{sec ( y )}{co tan ( y )} d y

2 x y y^{^{'}} = y^{2} - x^{2}

h \to 0 lim (- 2 x - h + 4)

\int z z^{2} - 5 d z

cos^{2} (5 t)

What is area f(x)=8x+2x^2-x^3,g(x)=0 ?
The answer to area f(x)=8x+2x^2-x^3,g(x)=0 is 148/3