What is area y1=x^3+4x^2-6x-9,y2=3x^2 ?

The answer to area y1=x^3+4x^2-6x-9,y2=3x^2 is 34.18141…

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area y1=x^3+4x^2-6x-9,y2=3x^2

area

y 1 = x^{3} + 4 x^{2} - 6 x - 9, y 2 = 3 x^{2}

34.18141 \dots

Solution steps

Isolate

y

for

y 1 = x^{3} + 4 x^{2} - 6 x - 9 : y = x^{3} + 4 x^{2} - 6 x - 9

Isolate

y

for

y 2 = 3 x^{2} : y = \frac{3 x ^{2}}{2}

Apply the area formula:

\int_{- 3.48091 \dots}^{2.17155 \dots} x^{3} + 4 x^{2} - 6 x - 9 - \frac{3 x ^{2}}{2} d x

Solve

\int_{- 3.48091 \dots}^{2.17155 \dots} x^{3} + 4 x^{2} - 6 x - 9 - \frac{3 x ^{2}}{2} d x : 34.18141 \dots

The area is:

= 34.18141 \dots

f (x) = e^{7 x^{2}}, at x = 2

\int y - y e^{z} d x

\int \frac{r}{( r ^{2} + x ^{2} ) ^{\frac{3}{2}}} d r

x \to 6 - lim (\frac{5}{x - 6})

\frac{\partial}{\partial y} (1 - x^{2})

What is area y1=x^3+4x^2-6x-9,y2=3x^2 ?
The answer to area y1=x^3+4x^2-6x-9,y2=3x^2 is 34.18141…