What is area 3x^2+8x+2,2x^2+4x-1 ?

The answer to area 3x^2+8x+2,2x^2+4x-1 is 4/3

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area 3x^2+8x+2,2x^2+4x-1

area

3 x^{2} + 8 x + 2, 2 x^{2} + 4 x - 1

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

Apply the area formula:

\int_{- 3}^{- 1} 3 x^{2} + 8 x + 2 - (2 x^{2} + 4 x - 1) d x

Solve

\int_{- 3}^{- 1} 3 x^{2} + 8 x + 2 - (2 x^{2} + 4 x - 1) d x : \frac{4}{3}

The area is:

= \frac{4}{3}

x^{2}, x + 2, x = - 2

\int \frac{5}{x ^{8}} d x

sin^{2} (x), 0

\int 10 sin^{2} (x) cos^{3} (x) d x

\frac{d}{d x} (ln (3 x))

What is area 3x^2+8x+2,2x^2+4x-1 ?
The answer to area 3x^2+8x+2,2x^2+4x-1 is 4/3