The integral of (x^6)(e^x7) I did x^7/7 U=e^x7 Where do I go from here?
K KEYWEST17 New member Joined Jan 19, 2011 Messages 46 Jan 19, 2011 #1 The integral of (x^6)(e^x7) I did x^7/7 U=e^x7 Where do I go from here?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jan 19, 2011 #2 Hello, KEYWEST17! Recall that: ∫eu du = eu+C\displaystyle \text{Recall that: }\;\int e^u\,du \:=\:e^u + CRecall that: ∫eudu=eu+C ∫x6ex7 dx\displaystyle \displaystyle \int x^6e^{x^7}\,dx∫x6ex7dx Click to expand... \(\displaystyle \text{We have: }\,\int e^{x^7}\left(x^6\,dx)\) Let: u = x7⇒du = 7x6 dx⇒x6 dx = 17 du\displaystyle \text{Let: }\:u \,=\,x^7 \quad\Rightarrow\quad du \,=\,7x^6\,dx \quad\Rightarrow\quad x^6\,dx \:=\:\tfrac{1}{7}\,duLet: u=x7⇒du=7x6dx⇒x6dx=71du Substitute: ∫eu(17 du) = 17∫eu du = 17eu+C\displaystyle \text{Substitute: }\;\int e^u\left(\tfrac{1}{7}\,du\right) \;=\;\tfrac{1}{7}\int e^u\,du \;=\;\tfrac{1}{7}e^u + CSubstitute: ∫eu(71du)=71∫eudu=71eu+C Back-substitute: 17ex7+C\displaystyle \text{Back-substitute: }\:\tfrac{1}{7}e^{x^7} + CBack-substitute: 71ex7+C
Hello, KEYWEST17! Recall that: ∫eu du = eu+C\displaystyle \text{Recall that: }\;\int e^u\,du \:=\:e^u + CRecall that: ∫eudu=eu+C ∫x6ex7 dx\displaystyle \displaystyle \int x^6e^{x^7}\,dx∫x6ex7dx Click to expand... \(\displaystyle \text{We have: }\,\int e^{x^7}\left(x^6\,dx)\) Let: u = x7⇒du = 7x6 dx⇒x6 dx = 17 du\displaystyle \text{Let: }\:u \,=\,x^7 \quad\Rightarrow\quad du \,=\,7x^6\,dx \quad\Rightarrow\quad x^6\,dx \:=\:\tfrac{1}{7}\,duLet: u=x7⇒du=7x6dx⇒x6dx=71du Substitute: ∫eu(17 du) = 17∫eu du = 17eu+C\displaystyle \text{Substitute: }\;\int e^u\left(\tfrac{1}{7}\,du\right) \;=\;\tfrac{1}{7}\int e^u\,du \;=\;\tfrac{1}{7}e^u + CSubstitute: ∫eu(71du)=71∫eudu=71eu+C Back-substitute: 17ex7+C\displaystyle \text{Back-substitute: }\:\tfrac{1}{7}e^{x^7} + CBack-substitute: 71ex7+C
