What is the integral from-1 to 3 of sqrt(x+1) ?

The integral from-1 to 3 of sqrt(x+1) is 16/3

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integral from-1 to 3 of sqrt(x+1)

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{4} u d u

= \int_{0}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\frac{\partial}{\partial y} (2 x^{3} y + 3)

\int_{- 3}^{- 5} - 4^{- 3} d x

\frac{d}{d x} (e^{- 16 x})

n = 1 \sum \infty \frac{6}{π ^{n}}

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

What is the integral from-1 to 3 of sqrt(x+1) ?
The integral from-1 to 3 of sqrt(x+1) is 16/3