Tom S. answered 05/27/20
Experienced, Patient Secondary School, College, and SAT/ACT Math Tutor
The area under the curve would be a definite integral. You want to do
∫15te-0.2tdt from 0 to 3. Use integration by parts.
I will start you out.
u = 15t and dv = e-0.2tdt Then du = 15dt and v = (1/-0.2)e-0.2t.
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