blackops23
Member
- Joined
- Dec 15, 2010
- Messages
- 428
- Gender
- Male
- HSC
- 2011
Hi, I just started integration a few days ago, and I'm up to the part of the Definite Integral and finding an area below the x-axis and all that stuff. Problem is, the Cambridge definition of definite integral with an integrand with negative values, is COMPLETELY different from the booklet my school teachers gave me (which includes explanations from J&C).
Basically here's the problem, I'm going to use an example to illustrate it.
f(x) = x^3 - 4x
Now I must integrate this in the bounds of x=2 and x=-2
As you can see for the graph, for -2(<)x(<)0, f(x) is positive, and for 0(<)x(<)2, f(x) is negative.
By the cambridge method, integration was done as normal, and the answer was 0.
By my booklet's method, the integration was split into two sections, one for the positive area and one for the negative area, except the absolute value of the negative area was taken.
So by cambridge's method, the answer was 4 - 4 =0, yet the booklet's answer was 4 - |- 4| = 8 units^2
So which one is right?
Basically here's the problem, I'm going to use an example to illustrate it.
f(x) = x^3 - 4x
Now I must integrate this in the bounds of x=2 and x=-2
As you can see for the graph, for -2(<)x(<)0, f(x) is positive, and for 0(<)x(<)2, f(x) is negative.
By the cambridge method, integration was done as normal, and the answer was 0.
By my booklet's method, the integration was split into two sections, one for the positive area and one for the negative area, except the absolute value of the negative area was taken.
So by cambridge's method, the answer was 4 - 4 =0, yet the booklet's answer was 4 - |- 4| = 8 units^2
So which one is right?
Last edited: