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Question: How do you evaluate the definite integral \[\int {\log xdx} \] from \(\left[ {2,4} \right]\)?...

How do you evaluate the definite integral ∫log⁡xdx\int {\log xdx} from [2,4]\left[ {2,4} \right]?

Explanation

Solution

In this question they have given ∫log⁡xdx\int {\log xdx} and asked us to solve it by using parts method. We will have to take any one term as uu and the other as dvdv and have to find its derivative and the value of vv. Then we need to substitute it in the formula I=uv−∫vduI = uv - \smallint vdu and then simply and solve it to find the correct answer.

Formula used: Formula for integrating in parts method:
I=uv−∫vduI = uv - \smallint vdu

Complete step by step solution:
In this question they have given ∫log⁡xdx\int {\log xdx} and asked us to solve it by using parts method.
We know that the formula for using part method is I=uv−∫vduI = uv - \smallint vdu
First we need to decide which one to choose as uu and which one as dvdv.
Let us take u=log⁡xu = \log x and dv=dx.dv = dx.
Now, we need to find the derivative of uuand the value of vv
⇒u=log⁡x\Rightarrow u = \log x
⇒dudx=ddxlog⁡x\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\log x
⇒dudx=1x\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{x}
With this we need to find dudu, which is
⇒du=1xdx\Rightarrow du = \dfrac{1}{x}dx
Now we know that dv=dx.dv = dx.
⇒dvdx=1\Rightarrow \dfrac{{dv}}{{dx}} = 1
⇒v=∫1dx\Rightarrow v = \int {1dx}
⇒v=x\Rightarrow v = x
Now, substituting in the formula, I=uv−∫vduI = uv - \smallint vdu
⇒∫log⁡xdx=x(log⁡x)−∫1x.xdx\Rightarrow \int {\log xdx} = x(\log x) - \smallint \dfrac{1}{x}.xdx
Sampling it,
⇒∫log⁡xdx=x(log⁡x)−∫dx\Rightarrow \int {\log xdx} = x(\log x) - \smallint dx
⇒∫log⁡xdx=x(log⁡x)−x+C\Rightarrow \int {\log xdx} = x(\log x) - x + C
Therefore, the integration of given II is∫(log⁡x)dx=xlog⁡x−x+C\int {(\log x)dx = } x\log x - x + C
Now applying the limit of
⇒∫24log⁡xdx=(4log⁡4−4)−(2(log⁡2)−2)\Rightarrow \int\limits_2^4 {\log xdx} = \left( {4\log 4 - 4} \right) - \left( {2(\log 2) - 2} \right)
⇒4log⁡22−4−2log⁡2+2\Rightarrow 4\log {2^2} - 4 - 2\log 2 + 2
⇒8log⁡2−2log⁡2−2\Rightarrow 8\log 2 - 2\log 2 - 2
Simplifying it, we get
⇒6log⁡2−2\Rightarrow 6\log 2 - 2

Therefore 6log⁡2−26\log 2 - 2 is the answer.

Note: Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. This method is used to find the integrals by reducing them into standard form. We do not add any constant while finding the integral of the second function.