What is the integral from 2.4 to 0.9 of 5/(6.4-10x) ?

The integral from 2.4 to 0.9 of 5/(6.4-10x) is 0.95619…

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integral from 2.4 to 0.9 of 5/(6.4-10x)

\int_{2.4}^{0.9} \frac{5}{6.4 - 10 x} d x

0.95619 \dots

Solution steps

\int_{2.4}^{0.9} \frac{5}{6.4 - 10 x} d x

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

= - \int_{0.9}^{2.4} \frac{5}{6.4 - 10 x} d x

Take the constant out:

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

= - 5 \cdot \int_{0.9}^{2.4} \frac{1}{6.4 - 10 x} d x

Apply u-substitution

: \int_{- 2.6}^{- 17.6} - \frac{1}{10 u} d u

= - 5 \cdot \int_{- 2.6}^{- 17.6} - \frac{1}{10 u} d u

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

= - 5 (- \int_{- 17.6}^{- 2.6} - \frac{1}{10 u} d u)

Take the constant out:

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

= - 5 (- (- \frac{1}{10} \cdot \int_{- 17.6}^{- 2.6} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 5 (- (- \frac{1}{10} [ln ∣ u ∣]_{- 17.6}^{- 2.6}))

Simplify

- 5 (- (- \frac{1}{10} [ln ∣ u ∣]_{- 17.6}^{- 2.6})) : - 0.5 [ln ∣ u ∣]_{- 17.6}^{- 2.6}

= - 0.5 [ln ∣ u ∣]_{- 17.6}^{- 2.6}

Compute the boundaries:

- 1.91238 \dots

= - 0.5 (- 1.91238 \dots)

Simplify

= 0.95619 \dots

\frac{x ^{2}}{sin ( x )}

x \to - 1 lim (\frac{x ^{5} + 1}{x + 1})

x \to 5 - lim (\frac{6 x}{x - 5})

\int \frac{1}{x ^{7}} - x^{7} - \frac{1}{2} d x

\int 2 y 2 y - 3 d y

What is the integral from 2.4 to 0.9 of 5/(6.4-10x) ?
The integral from 2.4 to 0.9 of 5/(6.4-10x) is 0.95619…