Certainly! Let's evaluate the integral:
[tex]\[
\int_{9}^{16} \frac{\ln y}{\sqrt{y}} \, dy
\][/tex]
To proceed, we'll break down the integration process step-by-step:
1. Substitution: We'll use the substitution [tex]\( y = u^2 \)[/tex]. Then, [tex]\( dy = 2u \, du \)[/tex].
When [tex]\( y = 9 \)[/tex]:
[tex]\[
u = \sqrt{9} = 3
\][/tex]
When [tex]\( y = 16 \)[/tex]:
[tex]\[
u = \sqrt{16} = 4
\][/tex]
2. Transform the integral: Substituting [tex]\( y \)[/tex] and [tex]\( dy \)[/tex] in the integral,
[tex]\[
\int_{9}^{16} \frac{\ln y}{\sqrt{y}} \, dy = \int_{3}^{4} \frac{\ln(u^2)}{\sqrt{u^2}} \cdot 2u \, du
\][/tex]
Simplifying the integrand:
[tex]\[
\int_{3}^{4} \frac{\ln(u^2)}{u} \cdot 2u \, du = \int_{3}^{4} \ln(u^2) \cdot 2 \, du
\][/tex]
[tex]\[
= 2 \int_{3}^{4} \ln(u^2) \, du
\][/tex]
3. Simplify logarithm: Use the property of logarithms [tex]\(\ln(u^2) = 2 \ln u\)[/tex],
[tex]\[
= 2 \int_{3}^{4} 2 \ln u \, du = 4 \int_{3}^{4} \ln u \, du
\][/tex]
4. Integration by parts: To integrate [tex]\( \ln u \)[/tex], we'll use integration by parts. Let [tex]\( v = \ln u \)[/tex] and [tex]\( dw = du \)[/tex]. Then [tex]\( dv = \frac{1}{u} du \)[/tex] and [tex]\( w = u \)[/tex].
Using the integration by parts formula [tex]\(\int v \, dw = vw - \int w \, dv \)[/tex]:
[tex]\[
\int \ln u \, du = u \ln u - \int u \left(\frac{1}{u}\right) du
\][/tex]
[tex]\[
= u \ln u - \int 1 \, du
\][/tex]
[tex]\[
= u \ln u - u
\][/tex]
So,
[tex]\[
\int_{3}^{4} \ln u \, du = \left[ u \ln u - u \right]_{3}^{4}
\][/tex]
5. Evaluate definite integral:
[tex]\[
= \left[ 4 \ln 4 - 4 \right] - \left[ 3 \ln 3 - 3 \right]
\][/tex]
[tex]\[
= (4 \ln 4 - 4) - (3 \ln 3 - 3)
\][/tex]
[tex]\[
= 4 \ln 4 - 4 - 3 \ln 3 + 3
\][/tex]
6. Combine and simplify:
[tex]\[
= 4 \ln 4 - 3 \ln 3 - 1
\][/tex]
7. Multiply the result by 4: Since we had the factor of 4 outside:
[tex]\[
4 \int_{3}^{4} \ln u \, du = 4 \left(4 \ln 4 - 3 \ln 3 - 1 \right)
\][/tex]
[tex]\[
= 16 \ln 4 - 12 \ln 3 - 4
\][/tex]
Now, evaluating the numerical value:
[tex]\[
16 \ln 4 - 12 \ln 3 - 4 \approx 4.997
\][/tex]
Thus, the evaluated integral is approximately:
[tex]\[
\int_{9}^{16} \frac{\ln y}{\sqrt{y}} \, dy \approx 4.997
\][/tex]
Matching this value, we determine that:
The integral [tex]\(\int_9^{16} \frac{\ln y}{\sqrt{y}} \, dy\)[/tex] evaluates to approximately 4.997.
