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?
[Math] Area under a log curve
integration
Related Solutions
When they say
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area under curve is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
They actually mean
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area between the curve and the $x$-axis is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
So when you're integrating you're only getting half the area:
and naturally, you need to multiply that by $2$ to get all of it.
Also, remember that the actual integrand in question here is not $y^2 = 16x$, because that's not of the form $y = f(x)$ as requested from the cases. It's actually $y = 4\sqrt{x}$, which is only the upper half of the parabola I've sketched above. That makes it a bit more obvious why you don't get all of it; when integrating $y = 4\sqrt x$, your expressions have no way of knowing that it is only half of something bigger, and it especially can't know that the other half is the exact mirror image. So it does the best it can and finds the area down to the $x$-axis instead.
Ok, so here's the cause of your confusion: the meaning of the word "area" depends on the context of the problem. Indeed your calculation
$$\int_{-2}^{2} (4x^3-16x)\,dx = 0$$
is correct. (A shortcut is to notice that you are integrating an odd function over a domain which is symmetric about the origin). However, in this particular question, the "area between the curves and the x-axis" really means to compute
$$\int_{-2}^{2} |4x^3-16x| \,dx$$
So yes, in the geometric sense, area generally must be positive. But when evaluating definite integrals, we sometimes think of the area above the x-axis as being "positive area" and the area below the x-axis as "negative area."
As to getting the answer to your problem, use symmetry to make your life easier:
$$\int_{-2}^{2} |4x^3-16x|\,dx = 2 \cdot \int_{-2}^{0} (4x^3-16x)\,dx = 2 \big[x^4 - 8x^2\big] \big|_{-2}^{0} = 32.$$
Best Answer
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.
$$5\ln(5) -4 \approx 4.05$$