What is the integral from 0 to 15 of sqrt(t+1) ?

The integral from 0 to 15 of sqrt(t+1) is 42

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integral from 0 to 15 of sqrt(t+1)

\int_{0}^{15} t + 1 d t

42

Solution steps

\int_{0}^{15} t + 1 d t

Apply u-substitution

: \int_{1}^{16} u d u

= \int_{1}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

42

= 42

\frac{d}{d x} ((1 + x^{3} Y^{2} Z)^{2})

\frac{d}{d x} (4^{x} ln (4))

y^{^{'}} (y - x^{2}) = - 2 y

\frac{\partial}{\partial x} (ln (x^{4} + y^{2}))

y^{^{'}} = 7 x + y, y (0) = 8

What is the integral from 0 to 15 of sqrt(t+1) ?
The integral from 0 to 15 of sqrt(t+1) is 42