C4 Integration Quiz (tough stuff)

Jun11

Written by Colin + in core 4, integration, quizzes.

C4 Integration quiz

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Can you integrate? You better had if you want to do well in C4.

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Question 1

How do you work out $\int \ln(x) dx$? (I want two methods.)

A	Parts ($u = 1,~v'=\ln(x)$) Hint: No, if you knew how to integrate $ln$, you wouldn't be in this mess.
B	You can only do it numerically Hint: Nut-uh. You can do it numerically, of course, but it's not the only way.
C	Substitution: $x = e^u$ Hint: Not a popular way, but a good way.
D	Parts ($u = \ln(x),~v'=1$) Hint: Yep, it drops out nicely.
E	You look it up - it's $\frac{1}{x} + C$ Hint: No, that's differentiating

Question 2

What method do you use to calculate $\int \sin^2(x) dx$? (Give me all four answers!)

A	Parts ($u = \sin(x),~v'=\sin(x)$) Hint: You can do! You need to do a trig identity in the next step, though.
B	Parts ($u = \sin^2(x),~v'=1$) Hint: Probably the most involved way - you have to do parts and a trig substitution in the second step.
C	Split-angle formula Hint: The way you get most help from the book with
D	None of the other four listed here. Hint: No! You mean any of the four listed here!
E	Trig substitution ($\cos(2x)$) Hint: Probably the easiest way.

Question 3

How do you work out $\int \tan^2(x) dx$?

A	Trig identity: $\tan^2(x) = \sec^2(x) - 1$ Hint: $sec^2(x)$ integrates to $\tan(x)$. It's in the book.
B	You can only do it numerically. Hint: Our survey said: ee-aw.
C	Substitution: start from $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$ Hint: It's a nice thought, but no. You end up with a $\frac{\tan(2x)}{\tan(x)}$ to integrate, which is no good at all.
D	Parts: $u = \tan^2(x),~v'=1$ Hint: This ends up as $x\tan^2(x) - \int 2x \tan^2(x)sec(x) dx$. I reckon it's possible, but I don't fancy it.
E	Parts: $u = \tan(x),~v'=\tan(x)$ Hint: Good luck with integrating $\sec^2(x) \ln(\sec(x)\tan(x))$ in the second step.

Question 4

What method do you use to calculate $\int \cos^9(x)\sin(x) dx$? (Give me both possibilities!)

A	Parts ($u = \sin(x),~v' =\cos^9(x)$) Hint: I don't know how to integrate $\cos^9(x)$, and neither do you! (Do you?)
B	Substitution ($u = \cos(x)$) Hint: Yep - my preferred way to tackle these.
C	Parts ($u = \cos^9(x),~v=\sin(x)$) Hint: In principle, that should work... eventually. It's a daft way to do it, though, so I'm marking it wrong.
D	Substitution ($u = \sin(x)$) Hint: That's not going to work.
E	Function-derivative Hint: Yes, $-\sin(x)$ is the derivative of $\cos(x)$, so you can use function-derivative.

Question 5

What method do you use to calculate $\int 2^x dx$?

A	Replace $2^x$ with $e^{x\ln(2)}$. Hint: Yep - messy, but it works.
B	Substitute $u = log_2(x)$ Hint: ... sorta. Differentiating $log_2(x)$ isn't trivial, though. I'm saying no.
C	Parts: $u = 2^x,~v'=1$ Hint: Good god, no. Have an ibuprofen.
D	Increase the power and divide by the new power. Hint: YOU JUST KILLED A KITTEN. YOU BASTARD!
E	You can only do it numerically Hint: Not true.

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Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

7 comments on “C4 Integration Quiz (tough stuff)”

MathbloggingAll
June 11, 2013 at 10:02 am

C4 Integration Quiz (tough stuff) http://t.co/NFQ24eRzbb

Reply →
A
April 20, 2015 at 2:57 pm

Good questions thanks

Maybe better if you show the answer for the actual maths at the end.

Just telling you but, isn’t the v=1 on the ln(x) question suppose to be v’=1? (2^x has one choice where it is v’=1)
Also, there is 80% chance that “non of the above” is not at the bottom so shouldn’t it be non of the others?

Reply →
- Colin
  April 20, 2015 at 9:32 pm
  
  Thanks for the feedback — I’ve shied away from giving the answers to the integrals, mainly out of laziness.
  
  Thanks especially for the corrections, I’ve fixed them 🙂
  
  Reply →
mkami wambura
December 30, 2016 at 6:40 pm

where are questions?????

Reply →
- Colin
  December 30, 2016 at 6:45 pm
  
  They should show up after you click the big blue ‘Start’ button.
  
  Reply →
aayush ganesh
April 14, 2019 at 8:20 pm

Differentiating log base 2 is actually fairly straightforward.

y = log2(x)

2^y = x

dx/dy = 2^y ln(2)

dy/dx = 1/((2^y)ln(2))

dy/dx = 1/xln(2)

Imo, this is easier than writing 2^x in terms of e^x and following the e^x integration by recognition rules.

The only reason it’s not used is because it’s not on the syllabus 🙁

They put differentiating and integrating any exponential but not any logarithm. Doesn’t make much sense to me.

Reply →
- Colin
  April 14, 2019 at 8:48 pm
  
  I’d probably do it as:
  
  $y = \log_2(x) = \frac{\ln(x)}{\ln(2)}$
  
  $\dydx = \frac{1}{x} \times \frac{1}{\ln(2)}$ directly.
  
  Reply →

C4 Integration quiz

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7 comments on “C4 Integration Quiz (tough stuff)”

A

Colin

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Colin

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Colin

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C4 Integration Quiz (tough stuff)

C4 Integration quiz

Colin

More from my site

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7 comments on “C4 Integration Quiz (tough stuff)”

MathbloggingAll

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Colin

mkami wambura

Colin

aayush ganesh

Colin

Leave a Reply Cancel reply

Where do you teach?

On twitter