Can Sketching Reveal the Solution to This Integral?

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SUMMARY

The integral $$\int_{1}^t 2\sin(t-\tau)e^{-2(t-1)} d\tau$$ can be simplified by factoring out constants and focusing on the integral $$\int_{1}^{t}\sin{(t-\tau)}d\tau$$. This approach allows for the evaluation of the area under the curve, leading to a function dependent solely on 't'. The discussion emphasizes the importance of sketching the constituent functions to visualize the solution, particularly in the context of convolution.

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wirefree
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I greatly appreciate this chance to submit a query.

I have the following integral: $$\int_{1}^t 2sin(t-\tau)e^{-2(t-1)} d\tau$$

and it has been suggested to me that if I sketch the two constituent functions and multiply them, I can read the answer off the paper. So here are my sketches: go straight to the 3rd arrow

1zme2rd.jpg


Don't let alternative integral side-track you; it involves something called "convolution" and is something I am grappling with too.


My firtst attempt: There is clearly some area under the graphs, so the answer is not 0. The alternatives I have are: 0, 1 or 2. But, surely, that's not something I can read that off the graph?

My second attempt: I just did the integral by taking $$2e^{-2(t-1)}$$ out; but that simply integrates down to an expression in terms of 't', and without any value of 't' I can't see how a definite answer can be obtained.


Any assistance with tackling the above would be helpful I am sure. Your advice is sought.wirefree
 
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I think you can put outside the integral the factors without ##\tau## so you can integrate ##\int_{1}^{t}\sin{(t-\tau)}d\,\tau## that is elementary, you obtain in this way an Area function that depends only by ##t##.
 

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