Area bounded by curve and axis
Curve and x-axis:
Area of shaded region =
$ \int_a^b f(x) \phantom{.} dx $
Area of shaded region =
$ - \int_a^b f(x) \phantom{.} dx = \int_b^a f(x) \phantom{.} dx \phantom{0}$
OR
$ \phantom{0} | \int_a^b f(x) \phantom{.} dx | \phantom{0}$
Curve and y-axis:
Area of shaded region =
$ \int_c^d g(y) \phantom{.} dy $
Area of shaded region =
$ - \int_c^d g(y) \phantom{.} dy = \int_d^c g(y) \phantom{.} dy \phantom{0}$
OR
$ \phantom{0} | \int_c^d g(y) \phantom{.} dy | \phantom{0}$
Area between two curves (or curve and line)
With respect to x-axis:
Area of shaded region =
$ \int_a^b f(x) \phantom{.} dx - \int_a^b g(x) \phantom{.} dx = \int_a^b f(x) - g(x) \phantom{.} dx $
With respect to x-axis:
Area of shaded region =
$ \int_c^d g(y) \phantom{.} dy - \int_c^d f(y) \phantom{.} dx = \int_c^d g(y) - f(y) \phantom{.} dx $
Question
1(a) On the same diagram, sketch the graphs of $(x - 1)^2 - (y - 1)^2 = 2$ and $y = (x - 0.5)^2 - 1$. Indicate clearly the axial intercepts, coordinates of point(s) of intersection and the equation of asymptotes (if any).
Solutions
\begin{align*}
(x - 1)^2 - (y - 1)^2 & = 2 \\
{(x - 1)^2 \over 2} - {(y - 1)^2 \over 2} & = 1 \\
{(x - 1)^2 \over (\sqrt{2})^2 } - {(y - 1)^2 \over (\sqrt{2})^2} & = 1 \\
\\
\text{(Left-Right) Hyperbola with } & \text{centre: } (1, 1) \\
\\
\text{Vertex: } (1 - \sqrt{2}, 1), & \phantom{0} (1 + \sqrt{2}, 1) \\
\\
\text{Let } {(x - 1)^2 \over 2} - {(y - 1)^2 \over 2} & = 0 \\
(x - 1)^2 - (y - 1)^2 & = 0 \\
(y - 1)^2 & = (x - 1)^2 \\
y - 1 & = \pm \sqrt{ (x - 1)^2 } \\
y - 1 & = \pm (x - 1) \\
\\
\text{Asymptotes: } y = x & \text{ or } y = - x + 2
\end{align*}
1(b) Find the area of the region by the curves $(x - 1)^2 - (y - 1)^2 = 2$ and $y = (x - 0.5)^2 - 1$.
Answer: $ 1.07 \text{ units}^2$
Solutions
\begin{align*}
(x - 1)^2 - (y - 1)^2 & = 2 \\
(x - 1)^2 & = 2 + (y - 1)^2 \\
x - 1 & = \pm \sqrt{2 + (y - 1)^2} \\
x & = 1 \pm \sqrt{2 + (y - 1)^2} \\
\\
y & = (x - 0.5)^2 - 1 \\
-(x - 0.5)^2 & = -1 - y \\
(x - 0.5)^2 & = y + 1 \\
x - 0.5 & = \pm \sqrt{y + 1} \\
x & = 0.5 \pm \sqrt{y + 1} \\
\\
\text{Area of region} & = \int_{y_1}^{y_2} x_1 - x_2 \phantom{.} dy \phantom{000000} [\text{'Right'} - \text{'Left'}] \\
& = \int_{0.2313}^{3.088} [1 - \sqrt{2 + (y - 1)^2} ] - [0.5 - \sqrt{y + 1} ] \phantom{.} dy \phantom{000000} [\text{Use GC}] \\
& = 1.0749 \\
& \approx 1.07 \text{ units}^2
\end{align*}
Definite integrals of modulus function:
Steps:
- Graph the modulus function and the original function (without modulus)
- For intervals where original function is positive, then $ \int f(x) \phantom{.} dx $
- For intervals where original function is negative, then $ - \int f(x) \phantom{.} dx $
2. Evaluate $ \int_{-1}^a |(x - 2)^2 - 4 \phantom{.} dx $, where $ 0 < a < 4 $, leaving your answer in terms of a.
Answer: $ -{1 \over 3} - {1 \over 3}(a - 2)^3 + 4a $
Solutions
\begin{align*}
\int_{-1}^a |(x - 2)^2 - 4| \phantom{.} dx & = \int_{-1}^0 (x - 2)^2 - 4 \phantom{.} dx + \left[ - \int_0^a (x - 2)^2 - 4 \phantom{.} dx \right] \\
\\
\text{Consider } \int (x - 2)^2 - 4 \phantom{.} dx & = \underbrace{\int (1) (x - 2)^2 \phantom{.} dx}_{ \int f'(x) [f(x)]^n \phantom{.} dx = {[f(x)]^{n + 1} \over n + 1}} - \int 4 \phantom{.} dx \\
& = { (x - 2)^3 \over 3 } - 4x \\
\\
\therefore \int_{-1}^a |(x - 2)^2 - 4| \phantom{.} dx & = \left[ { (x - 2)^3 \over 3 } - 4x \right]_{-1}^0 - \left[ { (x - 2)^3 \over 3 } - 4x \right]_0^a \\
& = \left[ {(0 - 2)^3 \over 3} - 4(0) \right] - \left[ {(-1 - 2)^3 \over 3} - 4(-1) \right] - \left[ { (a - 2)^3 \over 3} - 4a \right] + \left[ {(0 - 2)^3 \over 3} - 4(0) \right] \\
& = -{8 \over 3} + 5 - {(a - 2)^3 \over 3} + 4a - {8 \over 3} \\
& = -{1 \over 3} - {1 \over 3}(a - 2)^3 + 4a
\end{align*}
Definite integrals of periodic function:
3. It is given that
$$ f(x) = \begin{cases} x, & \text{ for } 0 \le x < 2 \\ -0.5x^2 + 2x , & \text{ for } 2 \le x < 4 \end{cases} $$
and that $f(x) = f(x + 4)$ for all real values of $x$.
(a) Sketch the graph of $y = f(x)$ for $-6 \le x \le 7$.
Solutions
Steps:
- Press y =
- Press math > scroll to B: piecewise(
- Select 2 pieces
- Enter equation and graph
- Since $f(x) = f(x + 4)$, the graph repeats itself every $4$ units
Sketch:
(b) Find $ \int_{-2}^7 f(x) \phantom{.} dx $.
Answer: $ 13{1 \over 6} $
Solutions
\begin{align*}
\int_{-4}^7 f(x) \phantom{.} dx & = \underbrace{\int_{-4}^0 f(x) \phantom{.} dx + \int_0^4 f(x) \phantom{.} dx}_\text{Same} + \int_4^7 f(x) \phantom{.} dx \\
& = 2 \int_0^4 f(x) \phantom{.} dx + \int_0^3 f(x) \phantom{.} dx \\
& = 2 \left[ \int_0^2 f(x) \phantom{.} dx + \int_2^4 f(x) \phantom{.} dx \right] + \int_0^2 f(x) \phantom{.} dx + \int_2^3 f(x) \phantom{.} dx \\
& = 3 \int_0^2 f(x) \phantom{.} dx + 2 \int_2^4 f(x) \phantom{.} dx + \int_2^3 f(x) \phantom{.} dx \\
& = 3 \int_0^2 x \phantom{.} dx + 2 \int_2^4 -0.5x^2 + 2x \phantom{.} dx + \int_2^3 -0.5x^2 + 2x \phantom{.} dx \\
& = 3 \left[ {x^2 \over 2}\right]_0^2 + 2 \left[ {-0.5x^3 \over 3} + {2x^2 \over 2}\right]_2^4 + \left[ {-0.5x^3 \over 3} + {2x^2 \over 2}\right]_2^3 \\
& = 6 + {16 \over 3} + {11 \over 6} \\
& = 13 {1 \over 6}
\end{align*}