What is the integral from 0 to 4 of sqrt(x)-1 ?

The integral from 0 to 4 of sqrt(x)-1 is 4/3

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integral from 0 to 4 of sqrt(x)-1

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

\int_{0}^{4} x d x = \frac{16}{3}

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

= \frac{16}{3} - 4

Simplify

\frac{16}{3} - 4 : \frac{4}{3}

= \frac{4}{3}

y = 5 - x^{2}

\int (6 x + 5) 3 x^{2} + 5 x d x

t \to 0 lim (\frac{4 t ^{2} + 3 t + 2}{t ^{3} + 2 t - 6})

f (x) = - x^{2} + 3 x - 1

\int \frac{( e ^{2 x} )}{e ^{4 x} + 4} d x

What is the integral from 0 to 4 of sqrt(x)-1 ?
The integral from 0 to 4 of sqrt(x)-1 is 4/3