What is the integral from 1 to b of 3cx^2+1 ?

The integral from 1 to b of 3cx^2+1 is c(b^3-1)+b-1

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integral from 1 to b of 3cx^2+1

\int_{1}^{b} 3 c x^{2} + 1 d x

c (b^{3} - 1) + b - 1

Solution steps

\int_{1}^{b} 3 c x^{2} + 1 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}^{b} 3 c x^{2} d x + \int_{1}^{b} 1 d x

\int_{1}^{b} 3 c x^{2} d x = c (b^{3} - 1)

\int_{1}^{b} 1 d x = b - 1

= c (b^{3} - 1) + b - 1

\int_{1}^{8} \frac{2 ln ( x )}{x} d x

\int_{0}^{1} 1 - y d y

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

\int_{0}^{1} 1 - t d t

\int_{0}^{π} 1 sin (n x) d x

What is the integral from 1 to b of 3cx^2+1 ?
The integral from 1 to b of 3cx^2+1 is c(b^3-1)+b-1