What is the integral from 2 to 3 of (39)/(sqrt(3-x)) ?

The integral from 2 to 3 of (39)/(sqrt(3-x)) is 78

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integral from 2 to 3 of (39)/(sqrt(3-x))

\int_{2}^{3} \frac{39}{3 - x} d x

78

Solution steps

\int_{2}^{3} \frac{39}{3 - x} d x

Take the constant out:

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

= 39 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 39 \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

= 39 (- \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

= 39 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 39 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

39 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 39 [2 u]_{0}^{1}

= 39 [2 u]_{0}^{1}

Compute the boundaries:

2

= 39 \cdot 2

Simplify

= 78

\int (- 3 x) d x

\int \frac{1}{2} x^{3} + 2 - (x + 1) d x

\frac{\partial}{\partial x} (2 x^{2} - 3 y - 4)

\frac{d}{d x} (e^{2 x} - 3)

(9 x^{2} - 15 x) e^{x}

What is the integral from 2 to 3 of (39)/(sqrt(3-x)) ?
The integral from 2 to 3 of (39)/(sqrt(3-x)) is 78