How do you work out $\int \ln(x) dx$? (I want two methods.)
A
You look it up - it's $\frac{1}{x} + C$
Hint:
No, that's differentiating
B
Parts ($u = \ln(x),~v'=1$)
Hint:
Yep, it drops out nicely.
C
Substitution: $x = e^u$
Hint:
Not a popular way, but a good way.
D
You can only do it numerically
Hint:
Nut-uh. You can do it numerically, of course, but it's not the only way.
E
Parts ($u = 1,~v'=\ln(x)$)
Hint:
No, if you knew how to integrate $ln$, you wouldn't be in this mess.
Question 5
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
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.
C
Substitution ($u = \cos(x)$)
Hint:
Yep - my preferred way to tackle these.
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.
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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.
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?
MathbloggingAll
C4 Integration Quiz (tough stuff) http://t.co/NFQ24eRzbb
A
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?
Colin
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 🙂
mkami wambura
where are questions?????
Colin
They should show up after you click the big blue ‘Start’ button.
aayush ganesh
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.
Colin
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.