Question

Sketch the graph of the given functions on the same set of axes, and shade the region bound by the two curves. Write down an expression that gives the area of the region. Find the area using a GDC.

$y = \frac{x + 4}{x - 2} and y = - x - 7$ .

Accepted Answer

Given functions: $y = \frac{x + 4}{x - 2} and y = - x - 7$

Now, find the $x -$ coordinates of the points of intersection of the two functions, i.e., $\frac{x + 4}{x - 2} = - x - 7$

$\Rightarrow - (x + 4) = (x + 7) (x - 2)$

$\Rightarrow - x - 4 = x^{2} - 2 x + 7 x - 14$

$\Rightarrow - x - 4 = x^{2} + 5 x - 14$

$\Rightarrow x^{2} + 6 x - 10 = 0$

$\Rightarrow x = \frac{- 6 \pm \sqrt{6^{2} - 4 \cdot 1 \cdot (- 10)}}{2}$

$\Rightarrow x = \frac{- 6 \pm \sqrt{76}}{2}$

$\Rightarrow x = \frac{- 6 \pm 8.72}{2}$ [Approximately]

$\Rightarrow x = \frac{- 6 + 8.72}{2}; \frac{- 6 - 8.72}{2}$ [Approximately]

$\Rightarrow x = 1.36; - 7.36$ [Approximately]

Now, the definite integral for the area between the two functions is $A = \int_{x_{1}}^{x_{2}} (g (x) - f (x)) d x$ i.e., $A = \int_{- 7.36}^{1.36} (\frac{x + 4}{x - 2} - (- x - 7)) d x$

Therefore, $A = \int_{- 7.36}^{1.36} (\frac{x - 2 + 6}{x - 2} - (- x - 7)) d x$

$\Rightarrow A = \int_{- 7.36}^{1.36} (1 + \frac{6}{x - 2} + x + 7) d x$

$\Rightarrow A = \int_{- 7.36}^{1.36} (8 + \frac{6}{x - 2} + x) d x$

$\Rightarrow A = {(8 x + 6 \ln |x - 2| + \frac{x^{2}}{2})}_{- 7.36}^{1.36}$

$\Rightarrow A = (8 (1.36) + 6 \ln |1.36 - 2| + \frac{{(1.36)}^{2}}{2}) - (8 (- 7.36) + 6 \ln |- 7.36 - 2| + \frac{{(- 7.36)}^{2}}{2})$

$\Rightarrow A = (10.88 + 6 \ln |- 0.64| + \frac{1.8496}{2}) - (- 58.88 + 6 \ln |- 9.36| + \frac{54.1696}{2})$

$\Rightarrow A = (11.8048 + 6 \ln (0.64)) - (- 31.7952 + 6 \ln (9.36))$

$\Rightarrow A \approx 27.5$ sq.units.

Hence, the area between two curves $A \approx 27.5$ sq.units.

Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: From Approximation to Generalization: Integration, Exercise 26: Exercise 10J

Natasha Awada Mathematics Solutions for Exercise - Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: From Approximation to Generalization: Integration, Exercise 26: Exercise 10J

Questions from Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: From Approximation to Generalization: Integration, Exercise 26: Exercise 10J with Hints & Solutions

Learn and Prepare for any exam you want !