What is the integral from-1 to 3 of-2x^2+4x+6 ?

The integral from-1 to 3 of-2x^2+4x+6 is 64/3

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integral from-1 to 3 of-2x^2+4x+6

\int_{- 1}^{3} - 2 x^{2} + 4 x + 6 d x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

\int_{- 1}^{3} - 2 x^{2} + 4 x + 6 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{3} 2 x^{2} d x + \int_{- 1}^{3} 4 x d x + \int_{- 1}^{3} 6 d x

\int_{- 1}^{3} 2 x^{2} d x = \frac{56}{3}

\int_{- 1}^{3} 4 x d x = 16

\int_{- 1}^{3} 6 d x = 24

= - \frac{56}{3} + 16 + 24

Simplify

- \frac{56}{3} + 16 + 24 : \frac{64}{3}

= \frac{64}{3}

\int \frac{cos ( x )}{1 - cos ^{2} ( x )} d x

- y^{^{'''}} - y^{^{''}} + 5 y^{^{''}} - 3 = 0

\frac{\partial}{\partial x} (17 - 4 x^{2} - y^{2})

\frac{\partial}{\partial x} (\frac{y}{cos ( x )})

\frac{d}{d x} (\frac{4}{5} ln ((9 x^{2} - 6)^{7}))

What is the integral from-1 to 3 of-2x^2+4x+6 ?
The integral from-1 to 3 of-2x^2+4x+6 is 64/3