Solve: \ \int_{0}^{ln(10)}\int_{e^{x}}^{10}\frac{1}{ln|y|}dy \ dx
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Nov 16, 2009 #1 Solve: ∫0ln(10)∫ex101ln∣y∣dy dx\displaystyle Solve: \ \int_{0}^{ln(10)}\int_{e^{x}}^{10}\frac{1}{ln|y|}dy \ dxSolve: ∫0ln(10)∫ex10ln∣y∣1dy dx
Solve: ∫0ln(10)∫ex101ln∣y∣dy dx\displaystyle Solve: \ \int_{0}^{ln(10)}\int_{e^{x}}^{10}\frac{1}{ln|y|}dy \ dxSolve: ∫0ln(10)∫ex10ln∣y∣1dy dx
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 16, 2009 #2 I see what is going on.No need for all that complicated Exponential log stuff. Use ddx∫h(x)g(x)f(t)dt=f(g(x))g′(x)−f(h(x))h′(x)\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)dxd∫h(x)g(x)f(t)dt=f(g(x))g′(x)−f(h(x))h′(x) and you will get 9. The fundamental theorem of calcarooney.
I see what is going on.No need for all that complicated Exponential log stuff. Use ddx∫h(x)g(x)f(t)dt=f(g(x))g′(x)−f(h(x))h′(x)\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)dxd∫h(x)g(x)f(t)dt=f(g(x))g′(x)−f(h(x))h′(x) and you will get 9. The fundamental theorem of calcarooney.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Nov 16, 2009 #3 galactus, you have to switch the order of integration, to wit:\displaystyle galactus, \ you \ have \ to \ switch \ the \ order \ of \ integration, \ to \ wit:galactus, you have to switch the order of integration, to wit: ∫0ln(10)∫ex101ln∣y∣dy dx = ∫110∫0ln∣y∣1ln∣y∣dx dy = 9.\displaystyle \int_{0}^{ln(10)}\int_{e^{x}}^{10}\frac{1}{ln|y|}dy \ dx \ = \ \int_{1}^{10}\int_{0}^{ln|y|}\frac{1}{ln|y|}dx \ dy \ = \ 9.∫0ln(10)∫ex10ln∣y∣1dy dx = ∫110∫0ln∣y∣ln∣y∣1dx dy = 9. Your answer is correct, usually people come up with the answer of 10.\displaystyle Your \ answer \ is \ correct, \ usually \ people \ come \ up \ with \ the \ answer \ of \ 10.Your answer is correct, usually people come up with the answer of 10. Good show.\displaystyle Good \ show.Good show.
galactus, you have to switch the order of integration, to wit:\displaystyle galactus, \ you \ have \ to \ switch \ the \ order \ of \ integration, \ to \ wit:galactus, you have to switch the order of integration, to wit: ∫0ln(10)∫ex101ln∣y∣dy dx = ∫110∫0ln∣y∣1ln∣y∣dx dy = 9.\displaystyle \int_{0}^{ln(10)}\int_{e^{x}}^{10}\frac{1}{ln|y|}dy \ dx \ = \ \int_{1}^{10}\int_{0}^{ln|y|}\frac{1}{ln|y|}dx \ dy \ = \ 9.∫0ln(10)∫ex10ln∣y∣1dy dx = ∫110∫0ln∣y∣ln∣y∣1dx dy = 9. Your answer is correct, usually people come up with the answer of 10.\displaystyle Your \ answer \ is \ correct, \ usually \ people \ come \ up \ with \ the \ answer \ of \ 10.Your answer is correct, usually people come up with the answer of 10. Good show.\displaystyle Good \ show.Good show.
