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QuestionSubstitute t=y+1t=y+1 in the integral ∫03yy+1dy\int_{0}^{3} \frac{y}{y+1} dy.

Studdy Solution

STEP 1

Assumptions1. We are given the integral ∫03yy+1dy\int_{0}^{3} \frac{y}{y+1} d y . We are asked to substitute t=y+1t=y+1 into the integral3. The limits of the integral are0 and3

STEP 2

First, we need to substitute t=y+1t=y+1 into the integral. This means we also need to change the differential dydy to dtdt and the limits of the integral.
∫0yy+1dy=∫14t−1tdt\int_{0}^{} \frac{y}{y+1} d y = \int_{1}^{4} \frac{t-1}{t} d t

STEP 3

Now, we can simplify the integrand by dividing each term in the numerator by the denominator.
∫1t−1tdt=∫1(1−1t)dt\int_{1}^{} \frac{t-1}{t} d t = \int_{1}^{} (1-\frac{1}{t}) d t

STEP 4

We can now split the integral into two separate integrals.
∫14(1−1t)dt=∫14dt−∫141tdt\int_{1}^{4} (1-\frac{1}{t}) d t = \int_{1}^{4} d t - \int_{1}^{4} \frac{1}{t} d t

STEP 5

The integral of1 with respect to t is just t, and the integral of 1t\frac{1}{t} with respect to t is ln⁡∣t∣\ln |t|. So, we can now compute the integrals.
∫14dt−∫141tdt=[t]14−[ln⁡∣t∣]14\int_{1}^{4} d t - \int_{1}^{4} \frac{1}{t} d t = [t]_{1}^{4} - [\ln |t|]_{1}^{4}

STEP 6

Substitute the limits of the integral into the expressions.
[t]14−[ln⁡∣t∣]14=(4−1)−(ln⁡∣4∣−ln⁡∣1∣)[t]_{1}^{4} - [\ln |t|]_{1}^{4} = (4-1) - (\ln |4| - \ln |1|)

STEP 7

implify the expressions.
(4−1)−(ln⁡∣4∣−ln⁡∣1∣)=3−(ln⁡4−0)=3−ln⁡4(4-1) - (\ln |4| - \ln |1|) =3 - (\ln4 -0) =3 - \ln4So, the result of the integral ∫03yy+1dy\int_{0}^{3} \frac{y}{y+1} d y after the substitution t=y+1t=y+1 is 3−ln⁡43 - \ln4.

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