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Can someone walk me through how to do the following problem so I can attempt a few more practice problems?

If: $$\int_{1}^{5} f(x) dx = 12 $$ and $$\int_{4}^{5} f( x) dx = 3.6$$ find: $$\int_{1}^{4} f( x) dx$$

Would it simply be $12 - 3.6$ ?


EDIT

If: $$\int_{0}^{9} f(x) dx = 37 $$ and $$\int_{0}^{9} g( x) dx = 16$$ find: $$\int_{0}^{9} 2f(x)+3g(x) dx$$

Would this simply be: $2 \times 37 + 3 \times 16$?

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Do you know any theorems or identities that relate the values of integrals of the same function with different limits? –  MJD May 7 '12 at 4:18
    
If $a \leq c \leq b$, then $\int_a^b=\int_a^c+\int_c^b$. Use this profitably. –  J. M. May 7 '12 at 4:19
    
As you said, it is simply 12-3.6. –  MJD May 7 '12 at 4:19
    
@MarkDominus Wow, I didn't think it would be that simple. That's why I posted it up here. I'll post a second one up so it's not a waste of a question. –  justcheckingin May 7 '12 at 4:21
    
Regarding the second problem: yes, that's right. –  Brian M. Scott May 7 '12 at 4:26

2 Answers 2

up vote 2 down vote accepted

This is done using additivity of integration on intervals i.e. if $c \in [a,b]$ and

$\displaystyle \int_a^b f(x) dx$, $\displaystyle \int_a^c f(x) dx$ and $\displaystyle \int_c^b f(x) dx$ are well- defined, then $$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$$ Hence, in your case, you have that $$\int_1^5 f(x) dx = \int_1^4 f(x) dx + \int_4^5 f(x) dx$$ Hence, we get that $$12 = \int_1^4 f(x) dx + 3.6$$ i.e. $$\int_1^4 f(x) dx = 8.4$$


For the second problem as Brian pointed out in the comments, it follows from linearity of integration i.e. $$\int \left( a f(x) + b g(x) \right)dx = a \int f(x)dx + b \int g(x) dx$$ provided all the integrals are well-defined.

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This is the additive property

$$\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx $$

The integral in the interval $[1,4]$ is the difference: integral in $[1,5]$ - integral $[4,5]$.

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