What is the integral from 0 to 2 of (30t^2)/(4+t^3) ?

The integral from 0 to 2 of (30t^2)/(4+t^3) is 10ln(3)

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integral from 0 to 2 of (30t^2)/(4+t^3)

\int_{0}^{2} \frac{30 t ^{2}}{4 + t ^{3}} d t

10 ln (3)

Decimal Notation

10.98612 \dots

Solution steps

\int_{0}^{2} \frac{30 t ^{2}}{4 + t ^{3}} d t

Take the constant out:

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

= 30 \cdot \int_{0}^{2} \frac{t ^{2}}{4 + t ^{3}} d t

Apply u-substitution

: \int_{4}^{12} \frac{1}{3 u} d u

= 30 \cdot \int_{4}^{12} \frac{1}{3 u} d u

Take the constant out:

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

= 30 \cdot \frac{1}{3} \cdot \int_{4}^{12} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 30 \cdot \frac{1}{3} [ln ∣ u ∣]_{4}^{12}

Simplify

30 \cdot \frac{1}{3} [ln ∣ u ∣]_{4}^{12} : 10 [ln ∣ u ∣]_{4}^{12}

= 10 [ln ∣ u ∣]_{4}^{12}

Compute the boundaries:

ln (12) - 2 ln (2)

= 10 (ln (12) - 2 ln (2))

Simplify

= 10 ln (3)

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What is the integral from 0 to 2 of (30t^2)/(4+t^3) ?
The integral from 0 to 2 of (30t^2)/(4+t^3) is 10ln(3)