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Question:

If $\displaystyle \int_{5.5}^{7} f(x)dx =-15$

then $\displaystyle \int_{7}^{5.5} (3 f(x)- 8)dx = ?$

My attempt:

A basic property of definite integrals says: $\displaystyle \int_{a}^{b} f(x)dx = \displaystyle -\int_{a}^{b} f(x)dx $

So therefore since $\displaystyle \int_{5.5}^{7} f(x)dx = -15$

$\displaystyle \int_{7}^{5.5} f(x)dx = 15$

Then multiply by $3$ and subtract $8$ and then we get $37$.

However, I know this is wrong because we're supposed to multiply $3$ and subtract $8$ from $f(x)$ and not $f(x)dx$. The problem is, I don't know how to separate $f(x)$ and $dx$ from each other .

3 Answers 3

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Use the linearity of the integral $$\begin{align}\int_7^{5.5}(3f(x)-8)\mathrm{d}x&=3\int_7^{5.5}f(x)\mathrm{d}x-8\int_7^{5.5}\mathrm{d}x\\ &= 3(15)-8(5.5-7) \\ &=57\end{align}$$ The linearity property of the integral means that for a constant $a$ and two functions $h(x)$ and $g(x)$ you have $$\int a h(x)\mathrm{d}x=a\int h(x)\mathrm{d}x$$ $$\int h(x)+g(x)\mathrm{d}x=\int h(x)\mathrm{d}x+\int g(x) \mathrm{d}x$$ If in the second property, you set $h(x)=3f(x)$ and $g(x)=-8$, you can use the first property to bring the $3$ out of the integral and use the identity you were given in the problem.

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    Why do you subtract 5.5-7? I think I understand everything else. You distribute the main expression and use the subtraction rule of integrals where you can separate it into 2 integrals. But why do you subtract 5.5-7? Does it have to do with only having dx and not f(x)dx? – 2017-01-18
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    @user6428243 it is just the evaluation of the second definite integral. $$\int_7^{5.5}1\mathrm{d}x=x|_7^{5.5}=5.5-7$$ – 2017-01-18
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We have $$I= 3\int_{7}^{5.5} f (x) dx -\int_{7}^{5.5} 8dx $$ $$=3 [-\int_{5.5}^{7} f (x) dx -8 \int_{7}^{5.5} dx] $$ Can you take it from here?

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You have $$\int_7^{5.5} (3f(x)-8)dx = 3\int_7^{5.5}f(x) - 8 \int_7^{5.5}dx=3\times 15 - 8 \times (-1.5) = 57.$$

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    My mistake, thanks you. – 2017-01-18