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

The integral from 0 to 4 of (x+1)^2 is 124/3

integral from 0 to 4 of (x+1)^2

\int_{0}^{4} (x + 1)^{2} d x

\frac{124}{3}

Decimal

41.33333 \dots

Solution steps

\int_{0}^{4} (x + 1)^{2} d x

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

= \int_{0}^{4} x^{2} + 2 x + 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_{0}^{4} x^{2} d x + \int_{0}^{4} 2 x d x + \int_{0}^{4} 1 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 2 x d x = 16

\int_{0}^{4} 1 d x = 4

= \frac{64}{3} + 16 + 4

Simplify

\frac{64}{3} + 16 + 4 : \frac{124}{3}

= \frac{124}{3}

d er i v a t i v e f (x) = - 4 x^{2} + 3 x

\frac{\partial}{\partial x} (e^{y} - sin (x))

\int_{0}^{2 π} \int 1 d x d y

\frac{\partial}{\partial z} (z^{3})

d er i v a t i v e sin (cos (x))

What is the integral from 0 to 4 of (x+1)^2 ?
The integral from 0 to 4 of (x+1)^2 is 124/3