Mixing
Integral of two functions: same interval and same variable
$begingroup$
Suppose I have the following two functions of x:
$f_1(x)=x^a*(z-x)^b$ and $f_2(x)=x^c*(z-x)^d$. Hence, only the exponents differ.
I found by trial and error that integrating $x$ out of both functions on the same interval results, and multiplying the result, equaled the integral of the product of functions. If this holds, I can sum exponents before doing the integration.
In short: the following appeared to happen:
$int_k^l f_1(x)dx int_k^l f_2(x)dx=int_k^l f_1(x)f_2(x)dx$.
I cannot find a theorem or rule to prove that is the case, although I was wondering whether I could use the sum rule if I log-transform $f_1(x)$ and $f_2(x)$, such that $int_k^l log(f_1(x))dx + int_k^l log(f_2(x))dx=int_k^l log(f_1(x))+log(f_2(x))dx$
1.) Is it valid to apply the sum rule here?
2.) Are there (other) theorems that apply in this situation?
integration
asked Feb 1 at 14:33
SuzanneSuzanne
11
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add a comment |
$begingroup$
Suppose I have the following two functions of x:
$f_1(x)=x^a*(z-x)^b$ and $f_2(x)=x^c*(z-x)^d$. Hence, only the exponents differ.
I found by trial and error that integrating $x$ out of both functions on the same interval results, and multiplying the result, equaled the integral of the product of functions. If this holds, I can sum exponents before doing the integration.
In short: the following appeared to happen:
$int_k^l f_1(x)dx int_k^l f_2(x)dx=int_k^l f_1(x)f_2(x)dx$.
I cannot find a theorem or rule to prove that is the case, although I was wondering whether I could use the sum rule if I log-transform $f_1(x)$ and $f_2(x)$, such that $int_k^l log(f_1(x))dx + int_k^l log(f_2(x))dx=int_k^l log(f_1(x))+log(f_2(x))dx$
1.) Is it valid to apply the sum rule here?
2.) Are there (other) theorems that apply in this situation?
integration
asked Feb 1 at 14:33
SuzanneSuzanne
11
$endgroup$
add a comment |
$begingroup$
Suppose I have the following two functions of x:
$f_1(x)=x^a*(z-x)^b$ and $f_2(x)=x^c*(z-x)^d$. Hence, only the exponents differ.
I found by trial and error that integrating $x$ out of both functions on the same interval results, and multiplying the result, equaled the integral of the product of functions. If this holds, I can sum exponents before doing the integration.
In short: the following appeared to happen:
$int_k^l f_1(x)dx int_k^l f_2(x)dx=int_k^l f_1(x)f_2(x)dx$.
I cannot find a theorem or rule to prove that is the case, although I was wondering whether I could use the sum rule if I log-transform $f_1(x)$ and $f_2(x)$, such that $int_k^l log(f_1(x))dx + int_k^l log(f_2(x))dx=int_k^l log(f_1(x))+log(f_2(x))dx$
1.) Is it valid to apply the sum rule here?
2.) Are there (other) theorems that apply in this situation?
integration
asked Feb 1 at 14:33
SuzanneSuzanne
11
$endgroup$
Suppose I have the following two functions of x:
$f_1(x)=x^a*(z-x)^b$ and $f_2(x)=x^c*(z-x)^d$. Hence, only the exponents differ.
I found by trial and error that integrating $x$ out of both functions on the same interval results, and multiplying the result, equaled the integral of the product of functions. If this holds, I can sum exponents before doing the integration.
In short: the following appeared to happen:
$int_k^l f_1(x)dx int_k^l f_2(x)dx=int_k^l f_1(x)f_2(x)dx$.
I cannot find a theorem or rule to prove that is the case, although I was wondering whether I could use the sum rule if I log-transform $f_1(x)$ and $f_2(x)$, such that $int_k^l log(f_1(x))dx + int_k^l log(f_2(x))dx=int_k^l log(f_1(x))+log(f_2(x))dx$
1.) Is it valid to apply the sum rule here?
2.) Are there (other) theorems that apply in this situation?
integration
integration
asked Feb 1 at 14:33
SuzanneSuzanne
11
asked Feb 1 at 14:33
SuzanneSuzanne
11
asked Feb 1 at 14:33
SuzanneSuzanne
11
asked Feb 1 at 14:33
SuzanneSuzanne
11
asked Feb 1 at 14:33
SuzanneSuzanne
11
11
add a comment |
add a comment |
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