Question

Find the area of the region bounded by the functions f(x) = x4 and g(x) =sq root IxI . (2 points)

7/15
14/15
28/15
None of these

Answer 1

I won't do your problem, but I will do a similar one.
\(\color{#b0000b}{ \displaystyle \small \bf\text{----------------------------------------------------------------} }\)
Find the area of the region bounded by \(\color{#000000}{ f(x)=\displaystyle x^2 }\)
and \(\color{#000000}{ \displaystyle g(x)=\sqrt{x~} }\).
\(\color{#b0000b}{ \displaystyle \small \bf\text{----------------------------------------------------------------} }\)
At first, the sketch of the region (shaded in black).
` https://www.desmos.com/calculator/rrfdhupsgh `
(copy paste this link to a new tab, and look at it)

You will see that your region starts from \(\color{#000000}{ x=0 }\)
and ends at \(\color{#000000}{ x=1 }\), and from there you know the
limits of integration.

Note, that if you were to just write
\(\color{#000000}{ \displaystyle\int\limits_0^1\sqrt{{\tiny~}x{\tiny~}}{\tiny~}dx }\)
then you are finding the following area:
` https://www.desmos.com/calculator/r2yztbqshj `
and that clearly exceeds the bounded region!

What if you tried to write
\(\color{#000000}{ \displaystyle\int\limits_0^1x^2{\tiny~}dx }\)
then you are finding the following area:
` https://www.desmos.com/calculator/zl97xehgig `
and that is just the area between the region
and the x-axis.

If you look at the graphs tho, you will not that
you can find the area of your region \(\Omega\), by
subtracting these two areas:
\(\color{#000000}{ \displaystyle\Omega=\left(\int\limits_0^1\sqrt{{\tiny~}x{\tiny~}}{\tiny~}dx\right)-\left(\int\limits_0^1x^2{\tiny~}dx\right) }\)
Look carefully at the graph below and make sure
you understand this observation:
` https://www.desmos.com/calculator/jubjyhaijj `

Now, after conceptualizing everything, we can
proceed to doing the math. The integrals (like
limits and derivatives) are linear (if they have
the same limits of integration), and that implies,
\(\color{#000000}{ \displaystyle\Omega=\int\limits_0^1\left(\sqrt{{\tiny~}x{\tiny~}}-x^2\right) {\tiny~}dx }\)

So, this is the integral you need to setup.
Now, we can apply the power rule term-by-term
and plug in the limits of integration.

\(\color{#000000}{ \displaystyle\Omega=\int\limits_0^1\left.\left(x^{1/2}-x^2\right) {\tiny~}dx=\left(\frac{2}{3}x^{3/2}-\frac{1}{3}x^3\right)\right|_{x=0}^{x=1} }\)

At \(x=0\) this is obviously 0, so we can write,
\(\color{#000000}{ \displaystyle\Omega=\left.\left(\frac{2}{3}x^{3/2}-\frac{1}{3}x^3\right)\right|^{x=1} =\frac{2}{3}-\frac{1}{3}=\color{red}{\frac{1}{3}} }\)