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

The integral from 1 to 4 of (sqrt(x)-1)^2 is 7/6

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

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

\frac{7}{6}

Decimal

1.16666 \dots

Solution steps

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

Expand

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

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

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 2 x d x = \frac{28}{3}

\int_{1}^{4} 1 d x = 3

= \frac{15}{2} - \frac{28}{3} + 3

Simplify

\frac{15}{2} - \frac{28}{3} + 3 : \frac{7}{6}

= \frac{7}{6}

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

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\int_{x}^{0} e^{- 3 t} d t

\int_{- 2}^{6} x^{3} d x

What is the integral from 1 to 4 of (sqrt(x)-1)^2 ?
The integral from 1 to 4 of (sqrt(x)-1)^2 is 7/6