how to evaluate this integral?

Question

I need to find the integral of T(t) = 20 + 75e^(-t/50) on the interval [0,30]. I've done it two ways and they don't give me consistent answers.

The first way I did it was to evaluate it as two separate integrals (20 and 75e^-t/50). I used substitution on the second integral so that it becomes -75e^u on the interval [0,-3/5]. This method gave me an answer of about 2292.

The second method I tried was to keep the two parts together. So when I did a substitution, it became -20 - 75e^u on the interval [0,-3/5]. This gave me an answer of about 46.

Which method is correct?? Or are both wrong?

Answer 1

Use the linear property of the integral, you can separate int[T(t)] into two separate integrals:

int[T(t)] = int(20 dt) + int(75e^[-t/50) dT). These are both simple integrals calculate. You get:


int[T(t)] = 20t + 75(-50)e^[-t/50] = 20t - (3750)e^[-t/50]

Now evaluate using your limits of integration [0->30]:

Upper limit (30): (20)(30) - (3750)e^[30/50] = 600-6862.9= -6232.9
Lower limit (0): (20)(0) - (3750)e^[0/50] = -3750

Thus, the answer is: -6238.9 + 3750 = -2488.9

My answer is slight off yours, so maybe it's just a miscalculation. Both methods are correct, but in the 2nd method, you need to factor at a 50

u = -t/50 ---> du = -1/50 dt, making dt = -50 du

If you compare the answers between method 2 and 1, you see that:

[50*method 2] approximately = method 1

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Hope this helps

Answer 2

Both methods are correct, but you have made a mistake in the second one.

If you let u = -t/50, then du/dt = -1/50. So when you make the substitution, you also have to replace dt by -50 du. Thus you'll get an extra factor of 50, and that will lead you back to the correct answer of 2292 or so.

(You must have done that correctly in the first substitution, so it must have just been a slip.)