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

The integral from 0 to 3 of 1/(sqrt(4-t)) is 2

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integral from 0 to 3 of 1/(sqrt(4-t))

\int_{0}^{3} \frac{1}{4 - t} d t

2

Solution steps

\int_{0}^{3} \frac{1}{4 - t} d t

Apply u-substitution

: \int_{4}^{1} - \frac{1}{u} d u

= \int_{4}^{1} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{4} - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{1}^{4} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{4}

= - (- [2 u^{\frac{1}{2}}]_{1}^{4})

Simplify

- (- [2 u^{\frac{1}{2}}]_{1}^{4}) : [2 u]_{1}^{4}

= [2 u]_{1}^{4}

Compute the boundaries:

2

= 2

\frac{\partial}{\partial x} (- z^{2})

f (x) = 5 x - 1

\int_{- 0.5}^{0} 1.5 x^{2} d x

x^{2} y^{^{''}} - 6 x y^{^{'}} + 10 y = 0

4 x^{3} - 4 x

What is the integral from 0 to 3 of 1/(sqrt(4-t)) ?
The integral from 0 to 3 of 1/(sqrt(4-t)) is 2