Thread: Define integral

The value of is

Originally Posted by xcelxp

The value of is

are you sure you have the limits right? usually they put the smaller number at the bottom. should the -2 be 2?

int{-2:0}[(x + 2)^2]dx

we proceed by substitution

let u = x + 2
=> du = dx

so our integral becomes

int{u^2}du
= (1/3)u^3 + C
= (1/3)(x + 2)^3 + C

evaluating between -2 and 0 we get
int{-2:0}[(x + 2)^2]dx = (1/3)(0 + 2)^3 - (1/3)(-2 + 2)^3
................................= 8/3

I am 100% sure that is the correct question. That doesnt seem to be the right answer.

The answer is -8/3.

Originally Posted by xcelxp

I am 100% sure that is the correct question. That doesnt seem to be the right answer.

i'm sure the integral is correct. switching around the limits to the order you used them just changes the answer to a negative value

int{0:-2}[(x + 2)^2]dx

we proceed by substitution

let u = x + 2
=> du = dx

so our integral becomes

int{u^2}du
= (1/3)u^3 + C
= (1/3)(x + 2)^3 + C

evaluating between 0 and -2 we get
int{0:-2}[(x + 2)^2]dx = (1/3)(-2 + 2)^3 - (1/3)(0 + 2)^3
................................= -8/3

In general:

Int[f(x) dx, b, a] = - Int[f(x) dx, a, b]

There is no logical restriction that a > b or b > a. It just depends on the kind of problem that you are doing. In Physics I have seen many instances where a > b.

-Dan