Math  /  Calculus

QuestionPart 2 of 3
Use the Substitution Formula, abf(g(x))g(x)dx=g(a)g(b)f(u)\int_{a}^{b} f(g(x)) \cdot g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) du where g(x)=ug(x)=u, to evaluate the following integral. 1102lnxxdx\int_{1}^{10} \frac{2 \ln x}{x} d x

Studdy Solution

STEP 1

What is this asking? We need to use the substitution formula to solve a definite integral with a logarithmic function inside. Watch out! Don't forget to change the limits of integration when you substitute!
Also, make sure you choose a *u* that simplifies the integral.

STEP 2

1. Define *u* and *du*
2. Substitute and solve
3. Calculate the boundaries

STEP 3

Let's **define** our *u*!
We want something that simplifies our integral, and we see that 1x\frac{1}{x} is the derivative of lnx\ln x.
So, let's set u=lnxu = \ln x.

STEP 4

Now, let's find *du*.
Taking the derivative of both sides, we get dudx=1x.\frac{du}{dx} = \frac{1}{x}. Multiplying both sides by dxdx gives us du=1xdxdu = \frac{1}{x} \cdot dx.
Perfect!

STEP 5

Let's **rewrite** our integral using our substitution: 1102lnxxdx=1102lnx1xdx=1102udu.\int_{1}^{10} \frac{2 \ln x}{x} dx = \int_{1}^{10} 2 \cdot \ln x \cdot \frac{1}{x} dx = \int_{1}^{10} 2 \cdot u \cdot du. See how much cleaner that looks?

STEP 6

Now, we can **integrate** with respect to *u*: 1102udu=2110udu=2[u22]110=[u2]110.\int_{1}^{10} 2 \cdot u \cdot du = 2 \int_{1}^{10} u \cdot du = 2 \cdot \left[ \frac{u^2}{2} \right]_{1}^{10} = \left[ u^2 \right]_{1}^{10}. We multiplied by 12\frac{1}{2} and then by 22, adding to one, to get the power of *u* from 11 to 22.

STEP 7

Remember, our original limits of integration were for *x*, not *u*.
We need to **convert** them!
Since u=lnxu = \ln x, when x=1x = 1, u=ln1=0u = \ln 1 = 0.
And when x=10x = 10, u=ln10u = \ln 10.

STEP 8

Now, we can **plug in** our *u* boundaries: [u2]110=[u2]0ln10=(ln10)2(0)2=(ln10)2.\left[ u^2 \right]_{1}^{10} = \left[ u^2 \right]_{0}^{\ln 10} = (\ln 10)^2 - (0)^2 = (\ln 10)^2.

STEP 9

Our **final answer** is (ln10)2(\ln 10)^2!

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