What is the integral from 0 to 1 of (a+bx^2) ?

The integral from 0 to 1 of (a+bx^2) is a+b/3

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integral from 0 to 1 of (a+bx^2)

\int_{0}^{1} (a + b x^{2}) d x

a + \frac{b}{3}

Solution steps

\int_{0}^{1} a + b x^{2} 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}^{1} a d x + \int_{0}^{1} b x^{2} d x

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

\int_{0}^{1} b x^{2} d x = \frac{b}{3}

= a + \frac{b}{3}

What is the integral from 0 to 1 of (a+bx^2) ?
The integral from 0 to 1 of (a+bx^2) is a+b/3