4. Which one gives the area of the region enclosed by the...

4. Which one gives the area of the region enclosed by the \\ 1\\curve $y = \frac{1}{x}$ and the lines $y = 2x$, $y = \frac{x}{4}$ ?\\(a) $\int_0^2 \frac{7}{2}xdx$\\(b) $\int_0^{\frac{1}{\sqrt{2}}} 7ydy + \int_{\frac{1}{\sqrt{2}}}^2 \frac{2 - y^2}{y}dy$\\(c) $\int_0^{\frac{1}{2}} 7ydy + \int_{\frac{1}{2}}^{\sqrt{2}} \frac{2 - y^2}{y}dy$\\(d) $\int_0^{\frac{1}{2}} \frac{7}{2}xdx + \int_{\frac{1}{2}}^{\sqrt{2}} \frac{4 - x^2}{2}dx$\\(e) $\int_0^{\frac{1}{\sqrt{2}}} 2xdx + \int_{\frac{1}{\sqrt{2}}}^2 \frac{4 - x^2}{4}dx$

Transcript

00:01 So in this question, i want to know what is the area of the region bounded by the curves, x equals negative y squared plus 13, and x equals y minus 1 quantity squared.

00:10 So my x equals negative y squared plus 13, i've gone ahead and graphed these on a computer on desmos.

00:19 Here is x equals negative y squared plus 13.

00:23 That's opening off to the left.

00:25 Whereas x equals y minus 1 quantity squared is this curve here.

00:31 So i'm going to have to figure out where these intersect in order to begin.

00:38 To figure that out, i'm going to set these equal to each other.

00:44 So negative y squared plus 13 equals y minus 1 quantity squared.

00:51 Now if i go ahead and foil out, multiply out the y minus 1 quantity squared, i'm getting y squared minus 2y plus 1.

01:05 Now i'm going to set this equal to 0.

01:07 To do that, i'm going to add over the y squared.

01:10 So that's 2y squared.

01:12 I've got my minus 2y.

01:14 I'm going to subtract over the 13.

01:18 So that's going to be minus 12.

01:21 I'll divide that right -hand side by 2.

01:25 So 0 equals y squared minus y minus 6.

01:30 Now that right -hand side, it factors.

01:33 It factors as the quantity of y minus 3 times the quantity of y plus 2, which gives us y values of 3 and negative 2.

01:47 And that makes sense based off of the graph that i see here.

01:50 This looks like y equals 3, whereas y equals negative 2 is down here.

01:57 So now what? well, we're going to be integrating with respect to y because i have functions of y here.

02:04 And i integrate in the order of right minus left.

02:08 So we're going to be integrating from y equals negative 2 to y equals 3.

02:15 And i'm integrating right minus left.

02:18 So negative y squared plus 13 minus my left function is y minus 1 being squared, all of this dy.

02:30 So let's see.

02:31 My antiderivative is negative y cubed over 3 plus 13y minus y minus 1 being cubed over 3...

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