What is the integral from 0 to 3 of (1+6x^2-10x^4) ?

The integral from 0 to 3 of (1+6x^2-10x^4) is -429

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integral from 0 to 3 of (1+6x^2-10x^4)

\int_{0}^{3} (1 + 6 x^{2} - 10 x^{4}) d x

- 429

Solution steps

\int_{0}^{3} 1 + 6 x^{2} - 10 x^{4} 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_{0}^{3} 1 d x + \int_{0}^{3} 6 x^{2} d x - \int_{0}^{3} 10 x^{4} d x

\int_{0}^{3} 1 d x = 3

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 10 x^{4} d x = 486

= 3 + 54 - 486

Simplify

= - 429

(1 + p x)^{- 1}

\frac{8}{ln ( x )}

sec^{2} (x)

\int \frac{cos ( 8 x )}{cos ^{2} ( 4 x )} d x

\int 1 + sin^{2} (x) d x

What is the integral from 0 to 3 of (1+6x^2-10x^4) ?
The integral from 0 to 3 of (1+6x^2-10x^4) is -429