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Integration help
- Thread starter Jaskaran
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BigGlenntheHeavy
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Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle \int\frac{3xdx}{\sqrt{4x+20}} \ = \ \frac{3}{2}\int\frac{xdx}{\sqrt{x+5}}\)
\(\displaystyle Now, \ let \ u \ = \ \sqrt{x+5}, \ du \ = \ \frac{dx}{2u} \ \implies \ dx \ = \ 2udu.\)
\(\displaystyle u \ = \ \sqrt{x+5}, \ \implies \ x \ = \ u^2-5.\)
\(\displaystyle Hence, \ we \ have: \ \frac{3}{2}\int\frac{(u^2-5)2udu}{u} \ = \ 3\int(u^2-5)du \ = \ 3[(u^3/3)-5u] \ + \ C,\)
\(\displaystyle = \ u^3-15u \ + \ C \ = \ (sub) \ (x+5)^{3/2}-15(x+5)^{1/2} \ + \ C \ = \ (x+5)^{1/2}[x+5-15]+ \ C,\)
\(\displaystyle = \ (x+5)^{1/2}(x-10)+C, \ I'll \ leave \ the \ check \ up \ to \ you.\)
\(\displaystyle \int\frac{3xdx}{\sqrt{4x+20}} \ = \ \frac{3}{2}\int\frac{xdx}{\sqrt{x+5}}\)
\(\displaystyle Now, \ let \ u \ = \ \sqrt{x+5}, \ du \ = \ \frac{dx}{2u} \ \implies \ dx \ = \ 2udu.\)
\(\displaystyle u \ = \ \sqrt{x+5}, \ \implies \ x \ = \ u^2-5.\)
\(\displaystyle Hence, \ we \ have: \ \frac{3}{2}\int\frac{(u^2-5)2udu}{u} \ = \ 3\int(u^2-5)du \ = \ 3[(u^3/3)-5u] \ + \ C,\)
\(\displaystyle = \ u^3-15u \ + \ C \ = \ (sub) \ (x+5)^{3/2}-15(x+5)^{1/2} \ + \ C \ = \ (x+5)^{1/2}[x+5-15]+ \ C,\)
\(\displaystyle = \ (x+5)^{1/2}(x-10)+C, \ I'll \ leave \ the \ check \ up \ to \ you.\)
BigGlenntheHeavy
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Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle \sqrt{4x+20} \ = \ \sqrt{4(x+5)} \ = \ \sqrt4 \sqrt{x+5} \ = \ 2\sqrt{x+5}\)
\(\displaystyle If \ you \ did \ not \ see \ this, \ then \ what \ can \ I \ say?\)
\(\displaystyle \sqrt{4x+20} \ = \ \sqrt{4(x+5)} \ = \ \sqrt4 \sqrt{x+5} \ = \ 2\sqrt{x+5}\)
\(\displaystyle If \ you \ did \ not \ see \ this, \ then \ what \ can \ I \ say?\)
Re: Integral of 3x/(Sqrt[4x+20])?
One last question, if I take the anti-derivative of this one more time, it's basically just expanding the factor and doing power rule, right?
And end up with Cx, right?
Ohhhh I see.BigGlenntheHeavy said:\(\displaystyle \sqrt{4x+20} \ = \ \sqrt{4(x+5)} \ = \ \sqrt4 \sqrt{x+5} \ = \ 2\sqrt{x+5}\)
\(\displaystyle If \ you \ did \ not \ see \ this, \ then \ what \ can \ I \ say?\)
One last question, if I take the anti-derivative of this one more time, it's basically just expanding the factor and doing power rule, right?
And end up with Cx, right?
BigGlenntheHeavy
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- Joined
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Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle Look, \ follow \ what \ I \ did \ above, \ as \ I'm \ not \ sure \ what \ you \ are \ talking \ about.\)
\(\displaystyle Look, \ follow \ what \ I \ did \ above, \ as \ I'm \ not \ sure \ what \ you \ are \ talking \ about.\)
Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle \If\ I \ wanted \ to \ take \ the \ integral \ of \int\sqrt{x+5}\\(x-10)\+\C\ one \ more \ time, \ what\ would \ it \ look \ like?\\)
? ? (x+5)*(x-10) , u = x-10, du=dx
?? u+C
=
I get 2(x-10)^(1/2) + Cx
\(\displaystyle \If\ I \ wanted \ to \ take \ the \ integral \ of \int\sqrt{x+5}\\(x-10)\+\C\ one \ more \ time, \ what\ would \ it \ look \ like?\\)
? ? (x+5)*(x-10) , u = x-10, du=dx
?? u+C
=
I get 2(x-10)^(1/2) + Cx
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle \int\sqrt{(x+5)} \ (x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5} \ + \ C\)
\(\displaystyle Use \ basically \ the \ same \ strategy.\)
\(\displaystyle \int\sqrt{(x+5)} \ (x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5} \ + \ C\)
\(\displaystyle Use \ basically \ the \ same \ strategy.\)
BigGlenntheHeavy
Senior Member
- Joined
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Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle OK \ Jaskaran, \ I \ get \ the \ impression \that \ you \ really \ want \ to \ learn \ "The \ Calculus",\)
\(\displaystyle so \ I \ am \ going \ to \ show \ you \ a \ few \ tips \ your \ teacher \ may \ have \ or \ may \ have \ not\)
\(\displaystyle shown \ you. \ First \ we \ want \ to \ find \ the \ anti-dervative \ of \ \int\sqrt{(x+5)} \ (x-10)dx, \ don't \ forget\)
\(\displaystyle the \ dx.\)
\(\displaystyle Now, \ a \ cute \ trick. \ Whenever \ you \ have \ a \ radical \ with \ index \ 2 \ (in \ other \ words \ the \ square \ root\)
\(\displaystyle of \ something), \ let \ u \ = \ \sqrt{(x+5)} \ = \ (x+5)^{1/2}.\)
\(\displaystyle This \ implies \ that \ du \ = \ \frac{1}{2}(x+5)^{-1/2}(1)dx \ = \ \frac{dx}{2(x+5)^{1/2}}.\)
\(\displaystyle However, \ (x+5)^{1/2} \ = \ u, \ so \ we \ have \ du \ = \ \frac{dx}{2u} \ or \ 2udu \ = \ dx, \ (we \ eliminated \ the\)
\(\displaystyle pesky \ radical).\)
\(\displaystyle Now, \ also \ note \ that \ u \ = \ (x+5)^{1/2} \ \implies \ that \ x \ = \ u^2-5.\)
\(\displaystyle Now, \ putting \ it \ all \ together, \ we \ get:\)
\(\displaystyle \int u(u^2-5-10)2udu \ = \ 2\int u^2(u^2-15)du \ = \ 2\int(u^4-15u^2)du \ = \ 2\bigg[\frac{u^5}{5}-5u^3\bigg]+C.\)
\(\displaystyle Now, \ resubbing, \ we \ get \ 2\bigg[\frac{(x+5)^{5/2}}{5}-5(x+5)^{3/2}\bigg]+C.\)
\(\displaystyle This \ gives \ \frac{2[(x+5)^{5/2}-25(x+5)^{3/2}]}{5}+C \ = \ \frac{2[(x+5)^{3/2}((x+5)-25)]}{5}+C.\)
\(\displaystyle Therefore, \ after \ all \ is \ said \ and \ done, \ we \ get: \ \int\sqrt{(x+5)}(x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5}+C.\)
\(\displaystyle Now, \ to \ elucidate \ furthur, \ how \ do \ we \ know \ we \ didn't \ mess \ up \ somewhere \ along \ the \ line?\)
\(\displaystyle Take \ a \ check, \ to \ wit: \ D_x\bigg[\frac{2[(x+5)^{3/2}(x-20)]}{5}+C\bigg] \ = \ \sqrt{x+5} (x-10), \ QED.\)
\(\displaystyle OK \ Jaskaran, \ I \ get \ the \ impression \that \ you \ really \ want \ to \ learn \ "The \ Calculus",\)
\(\displaystyle so \ I \ am \ going \ to \ show \ you \ a \ few \ tips \ your \ teacher \ may \ have \ or \ may \ have \ not\)
\(\displaystyle shown \ you. \ First \ we \ want \ to \ find \ the \ anti-dervative \ of \ \int\sqrt{(x+5)} \ (x-10)dx, \ don't \ forget\)
\(\displaystyle the \ dx.\)
\(\displaystyle Now, \ a \ cute \ trick. \ Whenever \ you \ have \ a \ radical \ with \ index \ 2 \ (in \ other \ words \ the \ square \ root\)
\(\displaystyle of \ something), \ let \ u \ = \ \sqrt{(x+5)} \ = \ (x+5)^{1/2}.\)
\(\displaystyle This \ implies \ that \ du \ = \ \frac{1}{2}(x+5)^{-1/2}(1)dx \ = \ \frac{dx}{2(x+5)^{1/2}}.\)
\(\displaystyle However, \ (x+5)^{1/2} \ = \ u, \ so \ we \ have \ du \ = \ \frac{dx}{2u} \ or \ 2udu \ = \ dx, \ (we \ eliminated \ the\)
\(\displaystyle pesky \ radical).\)
\(\displaystyle Now, \ also \ note \ that \ u \ = \ (x+5)^{1/2} \ \implies \ that \ x \ = \ u^2-5.\)
\(\displaystyle Now, \ putting \ it \ all \ together, \ we \ get:\)
\(\displaystyle \int u(u^2-5-10)2udu \ = \ 2\int u^2(u^2-15)du \ = \ 2\int(u^4-15u^2)du \ = \ 2\bigg[\frac{u^5}{5}-5u^3\bigg]+C.\)
\(\displaystyle Now, \ resubbing, \ we \ get \ 2\bigg[\frac{(x+5)^{5/2}}{5}-5(x+5)^{3/2}\bigg]+C.\)
\(\displaystyle This \ gives \ \frac{2[(x+5)^{5/2}-25(x+5)^{3/2}]}{5}+C \ = \ \frac{2[(x+5)^{3/2}((x+5)-25)]}{5}+C.\)
\(\displaystyle Therefore, \ after \ all \ is \ said \ and \ done, \ we \ get: \ \int\sqrt{(x+5)}(x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5}+C.\)
\(\displaystyle Now, \ to \ elucidate \ furthur, \ how \ do \ we \ know \ we \ didn't \ mess \ up \ somewhere \ along \ the \ line?\)
\(\displaystyle Take \ a \ check, \ to \ wit: \ D_x\bigg[\frac{2[(x+5)^{3/2}(x-20)]}{5}+C\bigg] \ = \ \sqrt{x+5} (x-10), \ QED.\)
D
Deleted member 4993
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Re: Integral of 3x/(Sqrt[4x+20])?
u = x+5
\(\displaystyle \int u^{\frac{1}{2}}(u-15)du \ = \ \int u^{\frac{3}{2}}du \ - \ 15*\int u^{\frac{1}{2}}du \ = \ \frac{2}{5}u^\frac{5}{2} \ - \ \ 10* u^\frac{3}{2} \ + \ C\)
and sub back.....
Jaskaran said:\(\displaystyle \If\ I \ wanted \ to \ take \ the \ integral \ of \int\sqrt{x+5}\\(x-10)\+\C\ one \ more \ time, \ what\ would \ it \ look \ like?\\)
? ? (x+5)*(x-10) , u = x-10, du=dx
?? u+C
=
I get 2(x-10)^(1/2) + Cx
u = x+5
\(\displaystyle \int u^{\frac{1}{2}}(u-15)du \ = \ \int u^{\frac{3}{2}}du \ - \ 15*\int u^{\frac{1}{2}}du \ = \ \frac{2}{5}u^\frac{5}{2} \ - \ \ 10* u^\frac{3}{2} \ + \ C\)
and sub back.....
BigGlenntheHeavy
Senior Member
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- 1,577
Re: Integral of 3x/(Sqrt[4x+20])?
\(\displaystyle Jaskaran, \ I'm \ losing \ patience \ with \ you.\)
\(\displaystyle Look \ at \ Subhotosh \ Khan \ work, \ it \ is \ probably \ easier \ to \ understand.\)
\(\displaystyle But \ if \ I \ started \ out \ with \ a \ C, \ what \ is \ that.\)
\(\displaystyle Jaskaran, \ I'm \ losing \ patience \ with \ you.\)
\(\displaystyle Look \ at \ Subhotosh \ Khan \ work, \ it \ is \ probably \ easier \ to \ understand.\)
\(\displaystyle But \ if \ I \ started \ out \ with \ a \ C, \ what \ is \ that.\)