What is the integral from 2 to 3 of (10)/(sqrt(3-x)) ?

The integral from 2 to 3 of (10)/(sqrt(3-x)) is 20

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integral from 2 to 3 of (10)/(sqrt(3-x))

\int_{2}^{3} \frac{10}{3 - x} d x

20

Solution steps

\int_{2}^{3} \frac{10}{3 - x} d x

Take the constant out:

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

= 10 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{u} d u

= 10 \cdot \int_{1}^{0} - \frac{1}{u} d u

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

= 10 (- \int_{0}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= 10 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{1}

= 10 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

10 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 10 [2 u]_{0}^{1}

= 10 [2 u]_{0}^{1}

Compute the boundaries:

2

= 10 \cdot 2

Simplify

= 20

\frac{d}{d x} (cos^{- 1} (x))

t^{2}

\int \frac{x + 7}{x} d x

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\frac{d}{d x} (π x^{7} - 2 x^{5} - 5 x^{- 2})

What is the integral from 2 to 3 of (10)/(sqrt(3-x)) ?
The integral from 2 to 3 of (10)/(sqrt(3-x)) is 20