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I'm given: $$\int_{5}^{-2} [3f(x) + 1]\,dx$$

with the additional information that: $$\int_{0}^{5} f(x)\,dx = 10$$

and $$\int_{0}^{-2} f(x)\,dx = -4$$

My layman mind looks at it as, since the sum of the two function pieces = 6, then the integral is $3(6) + 1 == 19$

Is that the right way of looking at the problem, or am I missing something?

Thanks for any assistance.

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+1 for explaining what you tried/what you were thinking. –  Eric Naslund Apr 7 '11 at 21:40

2 Answers 2

up vote 5 down vote accepted

That is nearly the right idea.

By linearity of the integral we have $$\int_{-2}^5 (3f(x)+1)dx=3\int_{-2}^5 f(x)dx+\int_{-2}^5 1 dx=3\int_{-2}^5 f(x)dx+7.$$ Notice the $+7$ instead of the $+1$, because we are integrating the constant function $1$ on an interval of length $7$. Also, it is true that $$\int_{-2}^5 f(x)dx =\int_{0}^5 f(x)dx+\int_{-2}^0 f(x)dx$$ but we are not given $\int_{-2}^0 f(x)dx$, instead we have $\int_{0}^{-2} f(x)dx$. How do these two integrals relate, and can you solve the problem from here?

Hope that helps,

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So now, I'm getting an answer of $49$ i.e. $3(14)+7$. That look right? –  NateyG Apr 7 '11 at 21:52
    
@NateyG: That definitely looks like the answer to me. Nice work. –  Eric Naslund Apr 7 '11 at 22:04

$$\int_5^{-2}(3f(x)+1)\,dx = 3\int_{5}^{-2}f(x)\,dx + \int_5^{-2}1\,dx.$$

But:

  1. $\int_5^{-2}1\,dx\neq 1$.

  2. $\int_5^{-2}f(x)\,dx \neq \int_0^5f(x)\,dx + \int_0^{-2}f(x)\,dx$.

Instead, remember that

  • $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$;
  • $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$;
  • $\int_a^b 1\,dx = b-a$.
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Typo: limits of integration (5 to -2) on the 1st integral on the RHS. –  Américo Tavares Apr 7 '11 at 21:55
    
@Américo: Thanks! Was away, so didn't see this. –  Arturo Magidin Apr 8 '11 at 2:48

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