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The curves
x=-(1)/(y^(2)-5) and
x=1 are graphed.
Which expression represents the area bounded by the curves?
Choose 1 answer:

The curves $x=-\frac{1}{y^{2}-5}$ and $x=1$ are graphed. $\newline$ Which expression represents the area bounded by the curves? $\newline$ Choose $1$ answer:

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Find derivatives using logarithmic differentiation

Full solution

Q. The curves

x=-\frac{1}{y^{2}-5}

and

x=1

are graphed.

\newline

Which expression represents the area bounded by the curves?

\newline

Choose

1

answer:

Set up integral: To find the area between the two curves, we need to set up an integral. The area $A$ can be found by integrating the difference between the two curves with respect to $y$ , from the lower intersection point to the upper intersection point. First, we need to find the points of intersection by setting the two expressions for $x$ equal to each other. $\newline$ $-\frac{1}{y^{2}-5} = 1$
Find intersection points: Now we solve for $y$ to find the points of intersection. Multiplying both sides by $-\left(y^2 - 5\right)$ to clear the fraction, we get: $\newline$ $1 = -\left(y^2 - 5\right)$
Solve for y: Next, we distribute the negative sign on the right side and bring all terms to one side to set the equation to zero: $\newline$ $0 = y^2 - 5 - 1$ $\newline$ $0 = y^2 - 6$
Set up integral for area: We solve for $y$ by taking the square root of both sides. Since we are looking for real intersection points, we consider both positive and negative roots: $\newline$ $y = \sqrt{6}$ and $y = -\sqrt{6}$ $\newline$ These are the $y$ -values of the points where the curves intersect.
Simplify integrand: With the intersection points found, we can set up the integral for the area $A$ as follows: $\newline$ $A = \int_{-\sqrt{6}}^{\sqrt{6}} \left(1 - \left(-\frac{1}{y^2 - 5}\right)\right) dy$ $\newline$ This integral represents the area under the curve $x=1$ minus the area under the curve $x=-\frac{1}{y^2-5}$ between the intersection points.
Integrate from $0$ to $\sqrt{6}$ : We simplify the integrand before integrating: $\newline$ $A = \int_{-\sqrt{6}}^{\sqrt{6}} \left(1 + \frac{1}{y^2 - 5}\right) dy$
Problem with integration: Now we integrate the function from $-\sqrt{6}$ to $\sqrt{6}$ . This integral is not elementary due to the term $(1)/(y^2 - 5)$ , and it requires partial fraction decomposition or a trigonometric substitution to solve. However, since the integrand is an even function (symmetric about the y-axis) and the limits of integration are symmetric about the y-axis, we can simplify the computation by integrating from $0$ to $\sqrt{6}$ and then doubling the result. $\newline$ $A = 2 \times \int_{0}^{\sqrt{6}} \left(1 + \frac{1}{y^2 - 5}\right) dy$
Problem with integration: Now we integrate the function from $-\sqrt{6}$ to $\sqrt{6}$ . This integral is not elementary due to the term $(1)/(y^2 - 5)$ , and it requires partial fraction decomposition or a trigonometric substitution to solve. However, since the integrand is an even function (symmetric about the y-axis) and the limits of integration are symmetric about the y-axis, we can simplify the computation by integrating from $0$ to $\sqrt{6}$ and then doubling the result. $\newline$ A = $2 \times \int_{0}^{\sqrt{6}} (1 + \frac{1}{y^2 - 5}) dy$ We integrate the function from $0$ to $\sqrt{6}$ . The integral of $1$ with respect to $y$ is simply $y$ . The integral of $\sqrt{6}$ $1$ requires a substitution or recognition of a standard integral form. However, this is where we encounter a problem: the integral of $\sqrt{6}$ $1$ is not elementary and cannot be expressed in terms of elementary functions. It involves an inverse hyperbolic function or a logarithmic function after a substitution. This indicates that we cannot proceed with the integration without using more advanced techniques, which are beyond the scope of this solution format. Therefore, we cannot provide a numerical answer for the area.

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