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The current is defined as follows:\$\$

\$ i=\begin{cases} 4 & 0<t<1 \\ 4t^2 & t > 1 \\ \end{cases} \$

The goal is to calculate the charge from \$t=0\$ to \$t=2\$ s

What I've tried:
I've tried integrating both parts.

\$\int 4dt=4t+c\$
\$\int 4t^2dt=\frac{4}{3}t^3+c\$

Then I fill in \$2\$ in the second integrated equation and \$0\$ in the first equation, and I subtract. I get \$\frac{32}{3}\$ coulombs as the answer, but I should get \$\frac{40}{3}\$ coulombs.

Where did I go wrong?

gilianzz
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2 Answers2

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You goofed the integration.

\$ \int_0^1 4dt + \int_1^2 4t^2dt \\ = \rvert_0^1 4t + \rvert_1^2{4\over3}t^3\\ = 4(1) - 4(0) + {4\over3}(2)^3 - {4\over3}(1)^3\\ = 4 + {28\over3}\\ = {40\over3}\$

Ignacio Vazquez-Abrams
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From 0 to 1: 12/3 C

From 1 to 2: 4/3*(8-1) = 28/3 C

12/3 C + 28/3 C= 40/3 C

Mario
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    In general, we try not to give direct answers to homework questions, but rather try to guide the OP toward finding the solution on his own. We're trying to teach the process, not the result. – Dave Tweed Sep 05 '16 at 21:14