Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A technician studies a circuit that has a rectifier in it. A rectifier lets the current pass in one direction only, so the charge keeps growing. The current at time seconds is amperes. The charge coulombs that has passed a point in the wire satisfies , and at no charge has passed, so . She notices that the power of is odd, so she keeps one factor for and writes the rest with . With the substitution she finds in the form . Find the constant , which is fixed by the condition at . Then find the charge that has passed at , and at . Give in coulombs and each charge in coulombs, to four decimal places.
Use and for even powers. For an odd power, peel off one factor and use .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. amperes and .
Plan. Peel off one , substitute , fix the constant, then evaluate.
Step 1. Substitute. , so the integral is . Thus .
Step 2. The constant. At , , so . Then .
Step 3. At . , so .
Step 4. At . . So .
Answer: , and the charges are and .
Check. The current is never negative for , so the charge grows. The values and then do grow.
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