I am trying to solve this
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I am confused on what the question is asking for. Which area do I need to solve for and what equation should I be using which leads me to an answer? Any help is appreciated. Thanks.
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2 Answers
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As calculus students, we know that for nice continuous functions like this
$$\begin{align} &\int_2^6 f'(x) \: \text{d}x = f(6)-f(2)\\\\\implies &f(2)+\int_2^6 f'(x) \: \text{d}x = f(6) \text{ .} \end{align} $$
I'm assuming you know to compute the integral above as "area under the curve."
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From the graph you provided, we know that (by area under curve method): $$\int^{6}_{2}f'(x)\cdot dx=4\pi$$ $$f(6)-f(2)=4\pi$$ $$f(6)=4\pi+3$$
