Area under a log curve

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?

• Can you also tell us what you have done? Or we cannot tell which parts go wrong. – user99914 May 10 '15 at 4:54
• As a quick check, the result ought to be $5\log5 - 4 \approx 4.05$. Did you evaluate your limits in the correct order? – suneater May 10 '15 at 4:59
• Could you show the antiderivative you obtained ? – Claude Leibovici May 10 '15 at 5:05
• 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. – Ian May 10 '15 at 5:10
• When they don't write the base, naturally it's the natural logarithm. – Alice Ryhl May 10 '15 at 5:47

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.

$$5\ln(5) -4 \approx 4.05$$

• Ah. Many thanks. – Ian May 10 '15 at 5:48

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.

• I think that even with the wrong limits (but in proper order) the result should be positive. – Claude Leibovici May 10 '15 at 5:08
• Yes, it is positive, even with the limits 0 and 4. – Chevy May 10 '15 at 5:14
• 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. – Ian May 10 '15 at 5:27
• You have to calculate the logarithm of base $e$, not $10$. – Chevy May 10 '15 at 5:33