• Apr 23rd 2007, 01:53 PM
bobchiba
Hi again, This could be covered in my exams and my text books don't cover it. Theres about 10 examples, if anyone could go thru this one I rekon I'll be good to do the rest.

• Apr 23rd 2007, 06:49 PM
Jhevon
Quote:

Originally Posted by bobchiba
Hi again, This could be covered in my exams and my text books don't cover it. Theres about 10 examples, if anyone could go thru this one I rekon I'll be good to do the rest.

the question is asking for the definate integral. do you know how to integrate?

here we have a nice polynomial, the integral is easy. we integrate the function term by term (i just have a flashback of power series, but that's not what i meant)

so anyway, we find the integral of powers of x in the following way:

int{x^n}dx = (x^{n+1})/(n + 1) + C for x not= -1

in other words, we add 1 to the power and divide by the new power. (if there is a constant in front, we just forget about it for a while and then multiply the function by it when done.)

for the definate integral, we take the formula we got for the indefinate integral, we plug in the upper limit (that is the number at the top of the integral sign) and we subtract the function when we plug in the lower limit. this is called the fundamental theorem of calculus, look it up, it's very important.

in math terms, if the int{f(x)}dx = F(x) then the int{from a up to b} f(x)dx = F(b) - F(a)

now let's do the problem

we have
f(x) = x^3 - 4x^2 + 3x

int{from 0 up to 1}f(x)dx = int{from 0 up to 1}(x^3 - 4x^2 + 3x)dx
..................................= [(1/4)x^4 - (4/3)x^3 + (3/2)x^2] evaluated between 0 and 1
..................................= [(1/4)(1)^4 - (4/3)(1)^3 + (3/2)(1)^2] - [(1/4)(0)^4 - (4/3)(0)^3 + (3/2)(0)^2]
..................................= [(1/4)(1)^4 - (4/3)(1)^3 + (3/2)(1)^2] - 0
..................................= 1/4 - 4/3 + 3/2
..................................= 5/12