Author Topic: PLease help (Read 5754 times) Tweet Share

khalil · « **Reply #15 on:** July 19, 2009, 11:19:34 pm »

What a great forum this turned out to be!

cobby · « **Reply #16 on:** July 20, 2009, 08:23:17 am »

$\int_a^b f(x) dx = f(b) - f(a)$

$\int f(x) dx = [x*ln(x) + (ln(2)-1)*x]_\frac{3}{2}^2$

Then sub your upper limit $(b=2)$ and lower limit $(a=3/2)$ and subtract from there to get your final answer.

$[x*ln(x) + (ln(2)-1)]^2 - [x*ln(x) + (ln(2)-1)*x]_\frac{3}{2}$

khalil · « **Reply #17 on:** July 20, 2009, 09:16:08 am »

Wait a second. How did you anti dervie log?

GerrySly · « **Reply #18 on:** July 20, 2009, 01:21:49 pm »

$\begin{align*} y &= x\cdot \log_e{(x)}\\ \frac{dy}{dx} &= 1+\log_e{(x)}\\ \therefore \int \log_e{(x)} &= \int \frac{dy}{dx}-1.dx\\ &=y-x\\ &=x\cdot \log_e{(x)} - x \end{align*}$

Now using that knowledge

$\begin{align*} \int_\frac{3}{2}^2 \log_e{(x)}.dx&=\left [ x\cdot \log_e{(x)} - x \right ]_\frac{3}{2}^2\\ &=\left [ (2) \cdot \log_e{(2)} - (2) \right ] - \left [ \left ( \frac{3}{2} \right ) \cdot \log_e{\left ( \frac{3}{2} \right ) - \left ( \frac{3}{2} \right ) } \right ]\\ &=2\cdot \log_e{(2)} - 2 - \frac{3}{2} \cdot \log_e{\left ( \frac{3}{2} \right ) } + \frac{3}{2}\\ &=\frac{1}{2}\left [ \log_e{(16)} - \log_e{ \left ( \frac{27}{8} \right )} \right ] - \frac{1}{2}\\ &=\frac{1}{2}\log_e{ \left ( \frac{128}{27} \right ) } - \frac{1}{2}\\ &=\frac{1}{2}\left [ \log_e{ \left ( \frac{128}{27} \right ) } - 1 \right ]\\ \end{align*}$

kamil9876 · « **Reply #19 on:** July 20, 2009, 01:31:34 pm »

Refer to attatched diagram

The way to do it without antidifferentiating log is to do it graphically. Notice how we can find the area we want by finding another area and then doing some subtraction etc. We can find the area by spltting the area in two, the rectangle at the bottom and the curvy part at the top. Finding the rectangle's area is easy, now all we need is a way to find the area of the curvy part at the top, and we can do that by integrating with respect to y:

$\int_c^d x dy$ This gives us the area trapped between the lines y=d, y=c, the y axis and the curve. To find that last curvy area we want we have to subtract the area we just found from the area of the big sideways rectangle:
$ Area_{curvy part}=2(d-c) - \int_c^d x dy$

Hence out total area is:
$ Area_{rectangle} + Area_{curvy part}= c(2 - \frac{3}{2}) + 2(d-c) - \int_c^d x dy$

shinny · « **Reply #20 on:** July 20, 2009, 04:54:46 pm »

Since it wasn't set out as a integration by recognition question, the way kamil did it is what they were expecting you to do. I think they had a very similar question on a methods exam a few years back which involved doing the exact same thing.

khalil · « **Reply #21 on:** July 22, 2009, 06:38:51 pm »

Thanks Kamil, but are you sure its possible to integrate along the y-axis?

khalil · « **Reply #22 on:** July 22, 2009, 06:40:12 pm »

Also, Find $\int_0^3 f(3x) dx$ if $\int_0^9 f(x) dx$ = 5

kamil9876 · « **Reply #23 on:** July 22, 2009, 07:02:10 pm »

Spec kids can just let u=3x here.

F(9) - F(0) = 5 where F is an antiderivative of f(x).

$\frac{d}{dx} F(3x)=3F'(3x)=3f(3x)$

$\therefore \int_0^3 3f(3x) dx=(F(3*3)-F(0))=(F(9)-F(0))=5$

Hence:

$\int_0^3 3f(3x) dx=5$
$\implies 3\int_0^3 f(3x) dx=5$

And the rest is trivial.

kamil9876 · « **Reply #24 on:** July 22, 2009, 07:05:49 pm »

Quote from: khalil on July 22, 2009, 06:38:51 pm

Thanks Kamil, but are you sure its possible to integrate along the y-axis?

Yes, it's like looking at the graph sideways sort of. (area doesn't change). I.e: just treat the y-axis as the x-axis, it's arbitrary which is x and y.

Although I do like your question (if it inspires a rigorous Cauchy-like proof involving limits and Riemman sums).

TrueTears · « **Reply #25 on:** July 22, 2009, 07:16:53 pm »

Quote from: khalil on July 22, 2009, 06:38:51 pm

Thanks Kamil, but are you sure its possible to integrate along the y-axis?

To integrate with respect to y is just using the inverse of your function which is in terms of x. It is in the methods course.

khalil · « **Reply #26 on:** July 22, 2009, 08:56:42 pm »

Quote from: kamil9876 on July 22, 2009, 07:02:10 pm

Spec kids can just let u=3x here.

F(9) - F(0) = 5 where F is an antiderivative of f(x).

$\frac{d}{dx} F(3x)=3F'(3x)=3f(3x)$

$\therefore \int_0^3 3f(3x) dx=(F(3*3)-F(0))=(F(9)-F(0))=5$

Hence:

$\int_0^3 3f(3x) dx=5$
$\implies 3\int_0^3 f(3x) dx=5$

And the rest is trivial.

I dont understand...are you stating that both integrals =5?

kamil9876 · « **Reply #27 on:** July 22, 2009, 09:00:32 pm »

which part specifically do you find unconvincing? Note the coefficient of 3 at the front, meaning the other integral is 5/3

khalil · « **Reply #28 on:** July 22, 2009, 09:49:45 pm »

Could you explain it differently....I dont understand the method you have administered! Sorry

kamil9876 · « **Reply #29 on:** July 22, 2009, 10:16:05 pm »

$\int_0^9 f(x) dx=F(9)-F(0)$ where F is an antiderivative of f(x).

using the value you are given for the first integral you can see:

$F(9)-F(0)=5$ (1)

Now notice that:

$\frac{d}{dx} (F(3x))=\frac{d(3x)}{dx} * \frac{F(3x)}{d(3x)}$ by the chain rule

$=3 * \frac{F(u)}{du}$ where u=3x

$=3f(u)$
$=3f(3x)$

Because $\frac{d}{dx}(F(3x))=3f(3x)$ we know that:

$\int_0^3 3f(3x) dx=[F(3x)]_0^3$
$=F(3*3)-F(0*3)$
$=F(9)-F(0)$ (2)

Subbing in equation (1) into (2) you get:

$\int_0^3 3f(3x) dx = 5$

But because of the wonderful property integrals that allows you to take out constant factors outside the integral you can change that equation by taking out the factor of 3:

$3\int_0^3 f(3x) dx = 5$

which gives $\int_0^3 f(3x)=\frac{5}{3}$

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