How do I evaluate Definite Integrals with negative index's? Integrals with negative index's?

Question

I do the problem: limit 2 over 1 3x^-2. When I anti differentiate I get 3(x^-1)/-1 which means -3x^-1. I substitute the limits into the problem and get: -3 times 2^-1 which equals 6. And: -3 times 1^-1 which equals 3. The difference is 3 which is my answer, but the book says that the answer is 1.5

Answer 1

> I do the problem: limit 2 over 1 3x^-2. When I anti differentiate I get 3(x^-1)/-1 which means -3x^-1.

That's right so far.

> I substitute the limits into the problem and get: -3 times 2^-1 which equals 6. And: -3 times 1^-1 which equals 3.

Nope. You forgot the ^-1, which means reciprocal. -3 times 2^-1 is actually -3 divided by 2, which is -1.5. 6 would've been -3 times 2. (-3/1 is still -3, though.)

> The difference is 3 which is my answer, but the book says that the answer is 1.5

The difference is -1.5 - (-3) = 1.5. Your integration was right, but your arithmetic was off (it happens to everyone).

Answer 2

Here is your problem:

-3*2^(-1) does not equal six, it equals -1.5. Similarly, -3*1^(-1) equals -3, not three. Thus the correct difference is -1.5-(-3), which is 1.5, which is the answer the book gave you.

Remember: x^(-y)=1/x^y

Answer 3

the integrated function is correct. your evaluation is wrong...

the evaulation is: -3(1/2) - (-3)(1)
=-3/2 +3
=1.5

remember, (x^-1) = 1/x

Answer 4

convinced this seems problematical to me besides. i imagine you want to to apply the trig identity cos 2x = a million - 2sin² x What i'd do is first convey f(x) as ½(a million - cos 2x) so as that f(x)g(x) will develop into ½(a million - cos 2x)e^x i'd now multiply out the brackets to provide ½e^x - ½e^xcos 2x Now i'd combine each and every expression yet recognising that i'd ought to apply Integration by using parts two times with the intention to combine the 2d expression.