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Let $\int\limits_4^7 f(x)\,dx = 2$, $\int\limits_6^7 f(x)\,dx = 17$, and $\int\limits_4^5 f(x)\,dx = 3$

Calculate $$\int\limits_5^6 f(x)\,dx$$

I guess I am to assume that $f(x)$ is the same in all integrals.

Should I figure out what $\Delta x$ is? Should I try to use identities of integrals?

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  • $\begingroup$ You're given information about $f$ (the three integrals), and you have to calculate what $\int_5^6 f(x) dx$ is. $\endgroup$ Apr 13, 2015 at 7:38

5 Answers 5

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${\int\limits_4^7}$ f(x) $dx = 2$

${\int\limits_4^7} f(x)dx = {\int\limits_4^5} f(x) dx+{\int\limits_5^6} f(x) dx + {\int\limits_6^7} f(x) dx $

Thus, ${\int\limits_5^6} f(x) dx = 2-17-3=-18$

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  • $\begingroup$ Why are you subtracting the numbers when you're summing up integrals? Confused here... $\endgroup$
    – dramadeur
    Apr 13, 2015 at 7:50
  • $\begingroup$ can you tell me which formula you use? I can't even google them out. Also, would they still apply if you had more than 3 integral terms? $\endgroup$
    – dramadeur
    Apr 13, 2015 at 7:53
  • $\begingroup$ To get to line three, @rightskewed has rearranged the equation in line two (check the limits on the integrals). The only "formula" used here is $\int_{a}^{b}f(x)\,\mathrm{d}x+\int_{b}^{c}f(x)\,\mathrm{d}x = \int_{a}^{c}f(x)\,\mathrm{d}x$ (which has technically been used a couple of times). $\endgroup$
    – Will R
    Dec 10, 2015 at 18:33
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$$\int_4^7 f(x) dx = \int_4^5 f(x) dx + \int_5^6 f(x) dx + \int_6^7 f(x) dx.$$ We're given three, and we have one unknown.

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$$ \int_4^5 f(x)\,dx + \int_5^6 f(x)\,dx + \int_6^7 f(x)\,dx = \int_4^7 f(x)\,dx $$

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  • $\begingroup$ Which formulas did you use? When I google "integral identities" I get some trigonometric stuff. $\endgroup$
    – dramadeur
    Apr 13, 2015 at 7:48
  • $\begingroup$ The above follows fairly straighforwardly from the definition of integrals. Consider the (somewhat simplfied) interpretation of integrals as "signed area under a curve", and it should be almost obvious. (See for example en.wikipedia.org/wiki/Integral#Conventions) $\endgroup$
    – mrf
    Apr 13, 2015 at 7:49
  • $\begingroup$ well, b and a changing (borders) isn't so obvious to me... Like how do you know that at the end (in this case) it's equal to integral which has b = 7 and a =4 $\endgroup$
    – dramadeur
    Apr 13, 2015 at 7:55
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    $\begingroup$ Think of it like this. Split the interval $[4,7]$ into three pieces. The total (signed) area under the curve is the sum of the areas corresponding to each interval respectively. $\endgroup$
    – mrf
    Apr 13, 2015 at 7:56
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Note: Although this problem is older the method displayed is useful as a secondary demonstration.

The problem presents three equations and asks to evaluate a fourth. Since the results are simple the form of the integrand can be considered in a simple form, ie $f(x) = a + b \, x + c \, x^2$. In this view it can be seen that: $$\int f(x) \, dx = a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}$$ for which \begin{align} \int_{4}^{7} f(x) \, dx = 2 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\ \int_{6}^{7} f(x) \, dx = 17 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\ \int_{4}^{5} f(x) \, dx = 3 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}. \end{align}

Solving this system f equations yields $$f(x) = \frac{1}{6} \, ( 4729 - 1806 \, x + 168 \, x^2)$$ and leads to $$\int_{5}^{6} f(x) \, dx = \frac{1}{6} \, [ 4729 \, x - 1403 \, x^2 + 56 \, x^3]_{5}^{6} = -18.$$

In comparison it is seen that $$\int_{5}^{6} f(x) \, dx = \left(\int_{4}^{7} - \int_{6}^{7} - \int_{4}^{5} \right) \, f(x) \, dx = (2 - 17 - 3) = -18.$$

An advantage for finding and solving the system of equations is by considering if the proposed question had asked for the value of the integral for the range (0,10), or (0,a), etc..

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You can apply this identity repetitively.
$ \int_a^b = \int_a^c+ \int_c^b $

$\int_4^7 f(x) dx = \int_4^5 f(x)dx + \int_5^7 f(x) dx$

$ \int_4^7 f(x) dx= \int_4^5 f(x)dx + \int_5^6 f(x) dx + \int_6^7 f(x) dx $

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