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

The answer to area x^2+2x+2,2x^2+3x-4 is 125/6

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

area

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

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 3}^{2} x^{2} + 2 x + 2 - (2 x^{2} + 3 x - 4) d x : \frac{125}{6}

The area is:

= \frac{125}{6}

\frac{d}{d x} (ln (x^{2} + 7))

x \to \frac{π}{2} lim (sin (x))

\int \frac{6}{x ^{2} + 10 x + 16} d x

\int 6 sin^{4} (x) d x

\frac{d}{d x} ((12 x^{2} - 12) (9 - 4 x))

What is area x^2+2x+2,2x^2+3x-4 ?
The answer to area x^2+2x+2,2x^2+3x-4 is 125/6