What is area y=2x^2,y=2x+6,-2,2 ?

The answer to area y=2x^2,y=2x+6,-2,2 is -(1+5sqrt(13))/3+6sqrt(13)+2

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area y=2x^2,y=2x+6,-2,2

area

y = 2 x^{2}, y = 2 x + 6, - 2, 2

- \frac{1 + 5 13}{3} + 613 + 2

Decimal Notation

17.29072 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{2} 2 x^{2} - (2 x + 6) d x

Solve

\int_{- 2}^{2} 2 x^{2} - (2 x + 6) d x : - \frac{1 + 5 13}{3} + 613 + 2

The area is:

= - \frac{1 + 5 13}{3} + 613 + 2

y \frac{d y}{d x} + 8 x^{3} = 4 x

\int tan (6 x) d x

\frac{d}{d x} (2 e^{x - 5})

y = - 9 x^{2}

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 0 ^{n}}

What is area y=2x^2,y=2x+6,-2,2 ?
The answer to area y=2x^2,y=2x+6,-2,2 is -(1+5sqrt(13))/3+6sqrt(13)+2