Question: User: https://www.youtube.com/watch?v=pJ0ckis1a6o
Transcript: hello everybody Welcome to my YouTube channel this is camba Jacob okay so I've got this question with me right over here and this is question 11 so we need to answer question a and question B so for question a we we evaluate the integral okay so you can see the integral we have it's uh the integral limits are from -1 to 2 and then here we've got 3 - 4x the^ 3 here so we are respecting X we shall say DX and this is how it looks so here the first thing that we're going to do will be to just say let U be equal to 3 - 4 x so whatever is inside here you let it be equal to U and then you say d u DX is equal to you find the derivative of whatever is here what is the derivative of three this is a constant so the derivative of three is a zero what is the derivative of -4x it will just be -4 that's a derivative of -4x now after getting this you make DX subject of the formula what number do we have down here it's a one so cross multiply du * 1 that will give us what a du you see du * 1 it's du equal to DX * -4 that is -4 DX you divide by4 4 remember we are making the x subject of the formula so this and this we go we shall remain with DX so this side we have negative du over A4 you see this is very important we need this part we need this one okay it's very important we also need this one so these two are the things we need so far so once we do this we go back now to the original question so we shall say integral okay equal to integral over u^ 3 D actually here it's a don't forget the negative du over a 4 so let me just explain what I've done here where there is a 3 - 4x I've just substituted the U u^ 3 is right here okay and where there is DX I've substituted there you see what is what we have here DX is here so this is what I've put here and this is how it looks let's clean up a little bit so to clean up here we have A1 over a four this negative can come this side and the four can come this outside the integral sign so we shall remain with the u^ 3 du so now from here we we of course integrate this so we should have a four here and a four here remember when integrating we always add 1 to the power so it would be 3 + 1 it's a 4 this power 4 you divide it right here and U we are going to have a -1/ 16 u^ 4 so -1 * u^ 4 it's what we have here and then 4 * 4 is 16 okay so now we can write back the value of U okay so remember remember we said let you be equal to what to 3 - 4 x and here power 4 don't forget the limits so we're going to put a line like this this is what we said we said let U be equal to this so we write it back and then the limits we've got uh negative uhga one here and then here it's a two these are the limits we have so now we can just substitute back the limits so here this is how it will look like then we have we have a 3 - 4 where there is X we put the upper limit which is a two then we close minus we write the same things inside here but where there's X we now put the lower limit which is A1 actually here there's a four you see this four also here there's a four okay so this is what we have now it's a matter of just simplifying okay you see so here uh we've got uh a 3 - 8 okay so when we multiply this times this we're getting a Min - 8 - 3 plus a 4 so when multiply -4 * -1 this is what we getting and then here we have this we have -5 this is -5 power 4 - 7 power 4 okay and when we work out further here this is what we getting what is uh what is uh what is -5^ 2 so I mean 5 power 4 Sorry 5^ 4 that is giving us a 625 minus here 7^ 4 it's uh giving us a 2,41 so we just simplify further we have -6 -1 / 16 MTI by negative uh -1 7 75 so when we do that this is giving us uh 111 so this is the answer we are getting 111 thank you so much for watching bye-bye
**Integral Evaluation of $(3 - 4x)^3$ from $-1$ to $2$**
**1. Problem Statement**
- Evaluate the integral $\int_{-1}^{2} (3 - 4x)^3 \, dx$
**2. Substitution Method**
- Let $u = 3 - 4x$
- Differentiate: $\frac{du}{dx} = -4$
- Rearrange for $dx$: $du = -4 dx \implies dx = -\frac{1}{4} du$
**3. Rewrite the Integral**
- Substitute $u$ and $dx$ into integral:
$\int (3 - 4x)^3 dx = \int u^3 \left(-\frac{1}{4} du\right) = -\frac{1}{4} \int u^3 du$
- Change the limits for $u$:
When $x = -1$, $u = 3 - 4(-1) = 7$
When $x = 2$, $u = 3 - 4(2) = -5$
- So integral becomes:
$-\frac{1}{4} \int_{7}^{-5} u^3 du = \frac{1}{4} \int_{-5}^{7} u^3 du$
**4. Integration and Evaluation**
- Integrate: $\int u^3 du = \frac{u^4}{4}$
- Apply limits:
$\frac{1}{4} \left[ \frac{u^4}{4} \right]_{-5}^{7} = \frac{1}{16} (7^4 - (-5)^4)$
- Calculate powers:
$7^4 = 2401$
$(-5)^4 = 625$
- Substitute values:
$\frac{1}{16} (2401 - 625) = \frac{1}{16} (1776) = 111$
**5. Final Answer**
- The value of the integral is $111$
**Summary:**
- Used substitution $u=3-4x$
- Transformed limits accordingly
- Performed integration of $u^3$
- Evaluated definite integral
- Result: $111$