why does integral 3/t^2 dt give u a negatrive answer?

Question

int 3/t^2 dt = -3/t but why -3 ?

Answer 1

Before integrating, the power of t is -2. After integrating, the power of t is -1, so the power rule indicates that we divide by -1.

As a response to the first answer, I agree that we should get in the habit of adding the constant to an indefinite integral, but your second point about a definite integral always being positive is not true.

If they specifically ask for the area between a curve and the x-axis, then it must be positive, and in fact you have to split up the integral into the pieces above the axis and the pieces below. But a definite integral that does not specifically represent area can have any sign.

Consider the definite integral of sin(x)dx from x = pi/2 to x = 2pi. That better come out negative because I'm taking an infinite sum of sin values, and there are a whole lot more negative sin values in that interval than positive ones. And it does come out negative; it's -1. If I wanted the area under the curve of sin(x) on that interval, I would have written |sin(x)|, the absolute value.

Answer 2

that is the indefinite integral

you are forgetting that there is always a constant of integration. if you perform a definite integration then you will always have a positive answer coz the area under the curve is always +ve

Answer 3

I = ∫ 3.t^(-2).dt
I = 3.t^(-1) / (-1) + C
I = - 3.t^(-1) + C
I = - 3 / t + C

Note 3 / (-1) = - 3

Answer 4

hmm..