Integrating an equation of constants raised to a common variable
up vote
1
down vote
favorite
Solving the integration problem:
$$int frac{4^x-5^x}{7^x} , dx$$
Not sure how to solve this problem, I don't see a way to incorporate integration by parts into this problem or even some kind of substitution.
Seems that taking the natural log of each element will help to simplify the equation but I don't think it is valid mathematically.
integration
edited Nov 14 at 17:33
Leucippus
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
add a comment |
up vote
1
down vote
favorite
Solving the integration problem:
$$int frac{4^x-5^x}{7^x} , dx$$
Not sure how to solve this problem, I don't see a way to incorporate integration by parts into this problem or even some kind of substitution.
Seems that taking the natural log of each element will help to simplify the equation but I don't think it is valid mathematically.
integration
edited Nov 14 at 17:33
Leucippus
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Solving the integration problem:
$$int frac{4^x-5^x}{7^x} , dx$$
Not sure how to solve this problem, I don't see a way to incorporate integration by parts into this problem or even some kind of substitution.
Seems that taking the natural log of each element will help to simplify the equation but I don't think it is valid mathematically.
integration
edited Nov 14 at 17:33
Leucippus
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
Solving the integration problem:
$$int frac{4^x-5^x}{7^x} , dx$$
Not sure how to solve this problem, I don't see a way to incorporate integration by parts into this problem or even some kind of substitution.
Seems that taking the natural log of each element will help to simplify the equation but I don't think it is valid mathematically.
integration
integration
edited Nov 14 at 17:33
Leucippus
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
edited Nov 14 at 17:33
Leucippus
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
edited Nov 14 at 17:33
Leucippus
19.5k102870
edited Nov 14 at 17:33
Leucippus
19.5k102870
edited Nov 14 at 17:33
Leucippus
19.5k102870
19.5k102870
asked Nov 14 at 16:33
YinGroen
62
asked Nov 14 at 16:33
YinGroen
62
asked Nov 14 at 16:33
YinGroen
62
62
Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42
add a comment |
Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42
Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42
add a comment |
2 Answers
2
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oldest
votes
up vote
3
down vote
HINT:
$$frac{4^x-5^x}{7^x}=left(frac47right)^x-left(frac57right)^x,$$
and you have the formula (by putting $a=4/7$ and $5/7$)
$$int a^x~mathrm dx=?$$
answered Nov 14 at 16:39
Tianlalu
2,594632
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
add a comment |
up vote
2
down vote
$$I=intfrac{a^x-b^x}{c^x}dx$$
$$I=intfrac{a^x}{c^x}dx-intfrac{b^x}{c^x}dx$$
$$I=int bigg(frac acbigg)^xdx-intbigg(frac bcbigg)^xdx$$
Each integral is in the form
$$G(k)=int k^xdx$$
$$G(k)=int e^{xlog k}dx$$
Substitute $u=xlog k$, which gives
$$G(k)=frac1{log k}int e^udu$$
$$G(k)=frac1{log k}e^{xlog k}$$
$$G(k)=frac1{log k}k^{x}+C$$
Thus
$$I=Gbigg(frac acbigg)-Gbigg(frac bcbigg)+C$$
Keep in mind $log(cdot)$ denotes the natural logarithm.
answered Nov 14 at 17:18
clathratus
1,841219
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
HINT:
$$frac{4^x-5^x}{7^x}=left(frac47right)^x-left(frac57right)^x,$$
and you have the formula (by putting $a=4/7$ and $5/7$)
$$int a^x~mathrm dx=?$$
answered Nov 14 at 16:39
Tianlalu
2,594632
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
add a comment |
up vote
3
down vote
HINT:
$$frac{4^x-5^x}{7^x}=left(frac47right)^x-left(frac57right)^x,$$
and you have the formula (by putting $a=4/7$ and $5/7$)
$$int a^x~mathrm dx=?$$
answered Nov 14 at 16:39
Tianlalu
2,594632
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
add a comment |
up vote
3
down vote
up vote
3
down vote
HINT:
$$frac{4^x-5^x}{7^x}=left(frac47right)^x-left(frac57right)^x,$$
and you have the formula (by putting $a=4/7$ and $5/7$)
$$int a^x~mathrm dx=?$$
answered Nov 14 at 16:39
Tianlalu
2,594632
HINT:
$$frac{4^x-5^x}{7^x}=left(frac47right)^x-left(frac57right)^x,$$
and you have the formula (by putting $a=4/7$ and $5/7$)
$$int a^x~mathrm dx=?$$
answered Nov 14 at 16:39
Tianlalu
2,594632
answered Nov 14 at 16:39
Tianlalu
2,594632
answered Nov 14 at 16:39
Tianlalu
2,594632
answered Nov 14 at 16:39
Tianlalu
2,594632
2,594632
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
add a comment |
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
@YinGroen: $$int a^x~mathrm dxne int x^a~mathrm dx.$$
– Tianlalu
Nov 14 at 17:05
add a comment |
up vote
2
down vote
$$I=intfrac{a^x-b^x}{c^x}dx$$
$$I=intfrac{a^x}{c^x}dx-intfrac{b^x}{c^x}dx$$
$$I=int bigg(frac acbigg)^xdx-intbigg(frac bcbigg)^xdx$$
Each integral is in the form
$$G(k)=int k^xdx$$
$$G(k)=int e^{xlog k}dx$$
Substitute $u=xlog k$, which gives
$$G(k)=frac1{log k}int e^udu$$
$$G(k)=frac1{log k}e^{xlog k}$$
$$G(k)=frac1{log k}k^{x}+C$$
Thus
$$I=Gbigg(frac acbigg)-Gbigg(frac bcbigg)+C$$
Keep in mind $log(cdot)$ denotes the natural logarithm.
answered Nov 14 at 17:18
clathratus
1,841219
add a comment |
up vote
2
down vote
$$I=intfrac{a^x-b^x}{c^x}dx$$
$$I=intfrac{a^x}{c^x}dx-intfrac{b^x}{c^x}dx$$
$$I=int bigg(frac acbigg)^xdx-intbigg(frac bcbigg)^xdx$$
Each integral is in the form
$$G(k)=int k^xdx$$
$$G(k)=int e^{xlog k}dx$$
Substitute $u=xlog k$, which gives
$$G(k)=frac1{log k}int e^udu$$
$$G(k)=frac1{log k}e^{xlog k}$$
$$G(k)=frac1{log k}k^{x}+C$$
Thus
$$I=Gbigg(frac acbigg)-Gbigg(frac bcbigg)+C$$
Keep in mind $log(cdot)$ denotes the natural logarithm.
answered Nov 14 at 17:18
clathratus
1,841219
add a comment |
up vote
2
down vote
up vote
2
down vote
$$I=intfrac{a^x-b^x}{c^x}dx$$
$$I=intfrac{a^x}{c^x}dx-intfrac{b^x}{c^x}dx$$
$$I=int bigg(frac acbigg)^xdx-intbigg(frac bcbigg)^xdx$$
Each integral is in the form
$$G(k)=int k^xdx$$
$$G(k)=int e^{xlog k}dx$$
Substitute $u=xlog k$, which gives
$$G(k)=frac1{log k}int e^udu$$
$$G(k)=frac1{log k}e^{xlog k}$$
$$G(k)=frac1{log k}k^{x}+C$$
Thus
$$I=Gbigg(frac acbigg)-Gbigg(frac bcbigg)+C$$
Keep in mind $log(cdot)$ denotes the natural logarithm.
answered Nov 14 at 17:18
clathratus
1,841219
$$I=intfrac{a^x-b^x}{c^x}dx$$
$$I=intfrac{a^x}{c^x}dx-intfrac{b^x}{c^x}dx$$
$$I=int bigg(frac acbigg)^xdx-intbigg(frac bcbigg)^xdx$$
Each integral is in the form
$$G(k)=int k^xdx$$
$$G(k)=int e^{xlog k}dx$$
Substitute $u=xlog k$, which gives
$$G(k)=frac1{log k}int e^udu$$
$$G(k)=frac1{log k}e^{xlog k}$$
$$G(k)=frac1{log k}k^{x}+C$$
Thus
$$I=Gbigg(frac acbigg)-Gbigg(frac bcbigg)+C$$
Keep in mind $log(cdot)$ denotes the natural logarithm.
answered Nov 14 at 17:18
clathratus
1,841219
answered Nov 14 at 17:18
clathratus
1,841219
answered Nov 14 at 17:18
clathratus
1,841219
answered Nov 14 at 17:18
clathratus
1,841219
1,841219
add a comment |
add a comment |
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Note that $d(a^x)/dx = a^x {rm ln} a$. This should set you on your way.
– Lucozade
Nov 14 at 16:39
Another hint is to note that $4^x = e^{x{rm ln} 4$.
– Lucozade
Nov 14 at 16:42