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

The integral from 1 to-1 of (x+1)^2 is -8/3

integral from 1 to-1 of (x+1)^2

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

- \frac{8}{3}

Decimal

- 2.66666 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} (x + 1)^{2} d x

Expand

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

= - \int_{- 1}^{1} 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_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} 2 x d x + \int_{- 1}^{1} 1 d x)

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

\int_{- 1}^{1} 2 x d x = 0

\int_{- 1}^{1} 1 d x = 2

= - (\frac{2}{3} + 0 + 2)

Simplify

- (\frac{2}{3} + 0 + 2) : - \frac{8}{3}

= - \frac{8}{3}

\int_{0}^{3} 1 + 4 r^{2} r d r

\int_{a}^{b} \frac{x ^{2}}{b - a} d x

\int_{\frac{π}{6}}^{\frac{π}{3}} 5 csc^{3} (x) d x

\int_{0}^{x} cosh^{2} (t) d t

\int_{0}^{x} \frac{1}{1 + t + t ^{2}} d t

What is the integral from 1 to-1 of (x+1)^2 ?
The integral from 1 to-1 of (x+1)^2 is -8/3