Trying to find area under log curve $f(x) = \log(1+x)$ with limits $x = 0$ to $4$. This part of the $\log$ curve is completely above the $x$ axis but when I integrate the function and apply the limits, the area always comes out negative. What am I doing wrong?
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1$\begingroup$ Can you also tell us what you have done? Or we cannot tell which parts go wrong. $\endgroup$ – user99914 May 10 '15 at 4:54
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$\begingroup$ As a quick check, the result ought to be $5\log5 - 4 \approx 4.05$. Did you evaluate your limits in the correct order? $\endgroup$ – suneater May 10 '15 at 4:59
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$\begingroup$ Could you show the antiderivative you obtained ? $\endgroup$ – Claude Leibovici May 10 '15 at 5:05
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$\begingroup$ Integral of log (x+1) = (x+1)log(x+1)-x and substituting in 4 and 0 gives me 5 x log5 - 4 = 5x.699-4 = -.505 - 0 = -.505. $\endgroup$ – Ian May 10 '15 at 5:10
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$\begingroup$ When they don't write the base, naturally it's the natural logarithm. $\endgroup$ – Alice Ryhl May 10 '15 at 5:47
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Ah, you're integrating $\log()$ as in the natural log, not base ten.
But you're calculating $\log_{10}()$.
$$5\log_{10}(5) -4 \approx -0.505$$
Try using $\ln()$ to process your calculation.
$$5\ln(5) -4 \approx 4.05$$
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I think you are integrating wrong. We have the integral $$\int_0^{4}\log (1+x)dx.$$ Let $t=1+x$. Then $dt=dx$ and $x_1=0, x_2=4$. We get $t_1=1$ and $t_2=5$. Thus $$\int_0^{4}\log (1+x)dx=\int_{1}^{5}\log t dt$$ Now, we use integration by parts: $u=\log t\Rightarrow du=dt/t$ and $dv=dt\Rightarrow v=t$. We have: $$\int_{1}^{5}\log t dt=t\log t|_{1}^{5}-\int_{1}^{5}dt=5\log 5-4$$ This is clearly $>0$.
I hope that helps. The indefinite integral of $\log (1+x)$ is $$\int \log (1+x) dx=-x+x\log (x+1)+\log (x+1)+ C,$$ where $C$ is a constant.
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$\begingroup$ I think that even with the wrong limits (but in proper order) the result should be positive. $\endgroup$ – Claude Leibovici May 10 '15 at 5:08
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$\begingroup$ Yes, it is positive, even with the limits 0 and 4. $\endgroup$ – Chevy May 10 '15 at 5:14
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$\begingroup$ Still can't get positive. 5 log5 = 5 x .699 = 3.495 - 4 = < 0. In the integration I used the 4 limit first and the 0 value second. $\endgroup$ – Ian May 10 '15 at 5:27
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$\begingroup$ You have to calculate the logarithm of base $e$, not $10$. $\endgroup$ – Chevy May 10 '15 at 5:33
