What is the integral from 0 to 1 of (3y^2)/(4+y^3) ?

The integral from 0 to 1 of (3y^2)/(4+y^3) is ln(5)-2ln(2)

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integral from 0 to 1 of (3y^2)/(4+y^3)

\int_{0}^{1} \frac{3 y ^{2}}{4 + y ^{3}} d y

ln (5) - 2 ln (2)

Decimal Notation

0.22314 \dots

Solution steps

\int_{0}^{1} \frac{3 y ^{2}}{4 + y ^{3}} d y

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{y ^{2}}{4 + y ^{3}} d y

Apply u-substitution

: \int_{4}^{5} \frac{1}{3 u} d u

= 3 \cdot \int_{4}^{5} \frac{1}{3 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{4}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{4}^{5}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{4}^{5} : [ln ∣ u ∣]_{4}^{5}

= [ln ∣ u ∣]_{4}^{5}

Compute the boundaries:

ln (5) - 2 ln (2)

= ln (5) - 2 ln (2)

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What is the integral from 0 to 1 of (3y^2)/(4+y^3) ?
The integral from 0 to 1 of (3y^2)/(4+y^3) is ln(5)-2ln(2)