What is the integral from 2 to 3 of (35)/(sqrt(3-x)) ?

The integral from 2 to 3 of (35)/(sqrt(3-x)) is 70

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integral from 2 to 3 of (35)/(sqrt(3-x))

\int_{2}^{3} \frac{35}{3 - x} d x

70

Solution steps

\int_{2}^{3} \frac{35}{3 - x} d x

Take the constant out:

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

= 35 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 35 \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

= 35 (- \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

= 35 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 35 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

35 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 35 [2 u]_{0}^{1}

= 35 [2 u]_{0}^{1}

Compute the boundaries:

2

= 35 \cdot 2

Simplify

= 70

(s^{6} + \frac{1}{s})^{\frac{5}{2}}

\frac{\partial}{\partial y} (3 x^{2} + 8 x y + y^{2})

\int \frac{2 x + 3}{2 x + 4} d x

p (x) = x^{3} - 5 x^{2} - 4 x + 20

8 + cos (t^{2})

What is the integral from 2 to 3 of (35)/(sqrt(3-x)) ?
The integral from 2 to 3 of (35)/(sqrt(3-x)) is 70