Question

Integrate from 4 to -2 ... |(x-1)(x-3)|dx

Answer 1

Did this question come with following choises

Answer 2

it did not

Answer 3

The instructions are simply to evaluate the integral

Answer 4

The integral of the absolute value is not a standard integral.

This is why they give us a definite integral. So we need to draw a picture.

(x-1)(x-3) is just a parabola whose zeros are at 1 and 3.
It also opens up, since the x^2 term is positive.
Therefore, the vertex must be in the middle of 1 and 3, at (2,-1).

However, we need to take the absolute value.
Our parabola will only be negative from x=1 to x=3, so we need to flip it over above the x-axis. See this picture: http://puu.sh/8oHTn.png

Now, we can integrate from 4 to -2. We just need to split it up. Do you know how to do that?

Answer 5

i think i have an idea but im better off saying no i dont know how. please help :)

Answer 6

First I want to clarify if the bounds are actually from 4 to -2? Or are they -2 to 4?
It's convention to read from the lower bound to the higher bound, so
\[ \int_a^b f(x) dx \] would read integrate f(x) from a to b.

Answer 7

correct. -2 to 4

Answer 8

Okay good.
So, the graph of the function is above in the link. Make sure you understood how I got to it.

From (-infinity, 1) our graph is just (x-1)(x-3). However, from (1,3) our graph is flipped because of the absolute value bars. From (3,infinity), it goes back to our original (x-1)(x-3).

So, we have three integrals now:

\[ \int_{-2}^1 (x-1)(x-3) \ dx + \int_{1}^3 -(x-1)(x-3) \ dx + \int_{3}^4 (x-1)(x-3) \ dx \]

The set-up is the difficult part, since this is the efficient way of evaluaing integrals of absolute values: to draw a picture.

Answer 9

excellent explanation. i totally understand so far

Answer 10

Check it out! WolframAlpha can't even express the indefinite integral in standard mathematical functions: http://puu.sh/8oIEg.png

The picture is necessary

Answer 11

lol yes. the picture helped a lot

Answer 12

well i am good. thank you very much

Answer 13

might you have time for another problem?

Answer 14

Sure.

Answer 15

integral 0 to ln3 of e^(x-e^x)

Answer 16

i assume that u substitution would come into play here. correct me if im wrong.

Answer 17

That's odd. This is the integral you just gave me?
\[ \int_0^{\ln(3)}e^{x-e^x} \ dx \]

Answer 18

yessir and yes i completely agree

Answer 19

with all respect, if it's not doable please refer me to some place where i might find the answer

Answer 20

You might have it solved here: http://math.stackexchange.com/

That's a very strange integral since it's only expressible in terms of the "exponential integral function." http://puu.sh/8oJyO.png

http://en.wikipedia.org/wiki/Exponential_integral

That's not an elementary function and it's odd that you have that as a question to solve. Perhaps your teacher/professor is just messing with you? lol

Answer 21

it's in my book called rogawski's calculus for AP and this is for a program called mathathon. my teacher chose problems from my book that would be a little more difficult than usual (mathathon questions were just too easy for an honors calculus class) to keep me busy during my week-long spring break

Answer 22

thanks for all the help

Answer 23

I have the 2nd edition of that book. What page/section can I find that on? I'm quite interested now.

Answer 24

hey turtlemuffin sorry i was on the website u told me about. its #48 in the chapter 5 review