Why isn’t the integral of $x^2$ from $0$ to $5$ equal to the integral of $\sqrt x$ from $0$ to $25$?

Question

My calculator says that the former is 41.67, and that the latter is 83.34. Why is that? Shouldn’t they be equal?

Answer 1

The integral of $x^2$ from $x=0$ to $x=5$ measures the area below the graph of $y=x^2$ from $x=0$ to $x=5$.

Now the inverse function $y=\sqrt{x}$ from $x=0$ to $x=25$ corresponds to reflecting the original function across the line $y=x$. So the integral from $x=0$ to $x=25$ of $\sqrt{x}$ is the area below the graph of $y=\sqrt{x}$ from $x=0$ to $x=5$, i.e., the area left of the graph of $y=x^2$ from $x=0$ to $x=5$.

So the areas are not equal, but their sum should be the area of the rectangle from $x=0$ to $x=5$ and $y=0$ to $y=25$, i.e., $125$.

Answer 2

Let me answer your question with a picture.

$$\int_0^5 x^2 \, dx$$ is the area of the red region while

$$\int_0^{25} \sqrt{y} \, dy$$

is the area of the blue region.

I have also constructed the line $y=5x$ to illustrate that they are not equal.