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

The integral from-1 to 4 of (x-3)^2+2 is 95/3

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

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

\frac{95}{3}

Decimal

31.66666 \dots

Solution steps

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

Simplify

(x - 3)^{2} + 2 : x^{2} - 6 x + 11

= \int_{- 1}^{4} x^{2} - 6 x + 11 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}^{4} x^{2} d x - \int_{- 1}^{4} 6 x d x + \int_{- 1}^{4} 11 d x

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

\int_{- 1}^{4} 6 x d x = 45

\int_{- 1}^{4} 11 d x = 55

= \frac{65}{3} - 45 + 55

Simplify

\frac{65}{3} - 45 + 55 : \frac{95}{3}

= \frac{95}{3}

What is the integral from-1 to 4 of (x-3)^2+2 ?
The integral from-1 to 4 of (x-3)^2+2 is 95/3