Proof for integral of sin(t)/t+1 dt from 0 to x

[p3forlife]

Homework Statement

Prove that the integral of sin(t)/t+1 dt from 0 to x is greater than 0 for all x > 0

Homework Equations

If f is bounded on [a,b], then f is integrable on [a,b] iff for every epsilon > 0 there exists a partition P of [a,b] s.t. U(f,P) - L(f,P) < epsilon.

The Attempt at a Solution

When you graph sin(t)/t+1 for t>=0, you get a sinusoidal graph with humps that get smaller and smaller, close to the horizontal axis. So using the definition of the integral as the area under the curve, it would make sense that integral of sin(t)/t+1 dt from x to 0 is greater than 0 along the positive x-axis, since the curve becomes almost insignificant as x>0, so the only area that "counts" is the first hump.

I'm stuck on how to do a formal proof, though. Any help would be much appreciated. Moreover, the graph is not bounded. Thanks!

 

[snipez90]

Hmmm, well the equation you have is irrelevant. Upper and Lower Sums are pretty useless here. A relevant equation might be the one that allows us to split a definite integral so that the [integral from a to b] is equal to the [integral from a to c] + [the integral from c to b].

Whenever you have a trig function as part of the integrand, you should consider symmetry, especially with sin and cos. This general idea usually works out pretty well. As for what the upper and lower limits of integration should be, I'll let you think about that more.