What is the integral from 0 to 16 of (500x)/(x^2+24) ?

The integral from 0 to 16 of (500x)/(x^2+24) is 250ln(35/3)

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integral from 0 to 16 of (500x)/(x^2+24)

\int_{0}^{16} \frac{500 x}{x ^{2} + 24} d x

250 ln (\frac{35}{3})

Solution steps

\int_{0}^{16} \frac{500 x}{x ^{2} + 24} d x

Take the constant out:

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

= 500 \cdot \int_{0}^{16} \frac{x}{x ^{2} + 24} d x

Apply u-substitution

: \int_{24}^{280} \frac{1}{2 u} d u

= 500 \cdot \int_{24}^{280} \frac{1}{2 u} d u

Take the constant out:

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

= 500 \cdot \frac{1}{2} \cdot \int_{24}^{280} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 500 \cdot \frac{1}{2} [ln ∣ u ∣]_{24}^{280}

Simplify

500 \cdot \frac{1}{2} [ln ∣ u ∣]_{24}^{280} : 250 [ln ∣ u ∣]_{24}^{280}

= 250 [ln ∣ u ∣]_{24}^{280}

Compute the boundaries:

ln (280) - ln (24)

= 250 (ln (280) - ln (24))

Simplify

= 250 ln (\frac{35}{3})

5 π x^{2}

\frac{\partial}{\partial x} (40 x^{\frac{3}{4}} y^{\frac{1}{4}})

y = x^{3}, x = - 1

f (x) = (x^{3} + 1) (x^{2} - 2), at x = 2

f (x) = (x - 4) (3 - x)^{3}, at x = 5

What is the integral from 0 to 16 of (500x)/(x^2+24) ?
The integral from 0 to 16 of (500x)/(x^2+24) is 250ln(35/3)