Area under a log curve

Question

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?

Answer 1

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.

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

Answer 2

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.