Evaluate the integral. 9 1 y ln(y) dy
The Correct Answer and Explanation is:
To evaluate the integral ∫19yln(y) dy\int_{1}^{9} y \ln(y) \, dy∫19yln(y)dy, we will use integration by parts.
Step 1: Identify parts for integration by parts
We will apply the formula for integration by parts:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu
In this case, we choose:
- u=ln(y)u = \ln(y)u=ln(y) (since the derivative of ln(y)\ln(y)ln(y) is simpler than yyy)
- dv=y dydv = y \, dydv=ydy
Now, differentiate and integrate:
- du=1y dydu = \frac{1}{y} \, dydu=y1dy
- v=y22v = \frac{y^2}{2}v=2y2
Step 2: Apply the integration by parts formula
Using the formula:∫19yln(y) dy=[y22ln(y)]19−∫19y22⋅1y dy\int_{1}^{9} y \ln(y) \, dy = \left[ \frac{y^2}{2} \ln(y) \right]_{1}^{9} – \int_{1}^{9} \frac{y^2}{2} \cdot \frac{1}{y} \, dy∫19yln(y)dy=[2y2ln(y)]19−∫192y2⋅y1dy
Simplify the second integral:∫19y22⋅1y dy=∫19y2 dy\int_{1}^{9} \frac{y^2}{2} \cdot \frac{1}{y} \, dy = \int_{1}^{9} \frac{y}{2} \, dy∫192y2⋅y1dy=∫192ydy
Step 3: Compute the first part
Now, evaluate the first term:[y22ln(y)]19=922ln(9)−122ln(1)\left[ \frac{y^2}{2} \ln(y) \right]_{1}^{9} = \frac{9^2}{2} \ln(9) – \frac{1^2}{2} \ln(1)[2y2ln(y)]19=292ln(9)−212ln(1)=812ln(9)−0= \frac{81}{2} \ln(9) – 0=281ln(9)−0
Since ln(1)=0\ln(1) = 0ln(1)=0, this term simplifies to:812ln(9)\frac{81}{2} \ln(9)281ln(9)
Step 4: Compute the second part
Now, evaluate the second integral:∫19y2 dy=12∫19y dy=12[y22]19\int_{1}^{9} \frac{y}{2} \, dy = \frac{1}{2} \int_{1}^{9} y \, dy = \frac{1}{2} \left[ \frac{y^2}{2} \right]_{1}^{9}∫192ydy=21∫19ydy=21[2y2]19=12(922−122)=12(812−12)= \frac{1}{2} \left( \frac{9^2}{2} – \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{81}{2} – \frac{1}{2} \right)=21(292−212)=21(281−21)=12×802=804=20= \frac{1}{2} \times \frac{80}{2} = \frac{80}{4} = 20=21×280=480=20
Step 5: Combine the results
Now combine the two parts:812ln(9)−20\frac{81}{2} \ln(9) – 20281ln(9)−20
Since ln(9)=2ln(3)\ln(9) = 2 \ln(3)ln(9)=2ln(3), we can rewrite the result as:812×2ln(3)−20=81ln(3)−20\frac{81}{2} \times 2 \ln(3) – 20 = 81 \ln(3) – 20281×2ln(3)−20=81ln(3)−20
Final Answer:
The value of the integral is:81ln(3)−2081 \ln(3) – 2081ln(3)−20
