Basic Mathematics · Unit 7
Applications of Integration
Exam-focused notes for Applications of Integration (Basic Mathematics, MTH104): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Area between curves
- Area under curves
- Volume of solids of revolution
- Disk and washer methods
- Shell method
- Arc length of curves
- Surface area of revolution
- Trapezoidal rule for numerical integration
Volume of solids of revolution
Find the volume of the solid obtained by rotating about the y-axis the region bounded by $y = x$ and $y = x^2$. [5]
- Curve 1: $y = x$ - Curve 2: $y = x^2$ - Axis of rotation: the y-axis - Region: bounded between the two curves All data needed is present. Step 1: Intersection points $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ So $x = 0$ or $x = 1$, giving inters...
Full solved answer →A pyramid 3 m3\text{ m}3 m high has a square base that is 3 m3\text{ m}3 m on a side. The cross section of the pyramid perpendicular to the altitude x mx\text{ m}x m down from the vertex is a square x mx\text{ m}x m on a side. Find the volume of the pyramid. [5]
- Height of pyramid: $h = 3$ m - Square base: $3$ m on a side - Cross section at distance $x$ m down from the vertex: a square that is $x$ m on a side - Cross-section side length as function of $x$: side $= x$ - Integration limits: from vertex $x = 0$ to ba...
Full solved answer →Area under curves
Find the area of the region between the x-axis and the graph of $f(x) = x^3 - x^2 - 2x$, $1 \leq x \leq 2$. [5]
- Function: $f(x) = x^3 - x^2 - 2x$ - Interval: $1 \leq x \leq 2$ - Find area between curve and x-axis. Step 1: Zeros of f(x) $$f(x) = x(x^2 - x - 2) = x(x-2)(x+1)$$ Zeros: $x = 0, 2, -1$. In $[1,2]$ the only zero is $x = 2$ (endpoint). Step 2: Sign of f(x)...
Full solved answer →Find the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Evaluate: $\int_0^2 \frac{x}{\sqrt{x^2 + 4}} , dx$.
[5+5]
Part 1: Ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ Part 2: Integral $\displaystyle\int0^2 \frac{x}{\sqrt{x^2 + 4}}\, dx$ --- (a) Area Enclosed by the Ellipse Compare with standard form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$: $$a^2 = 9 \Rightarrow a = ...
Full solved answer →Arc length of curves
Define arc-length of the curve. Find the length of the curve $y = \left(\frac{x}{2}\right)^{\frac{2}{3}}$ from $x = 0$ to $x = 2$. [1+4]
- Curve: $y = \left(\dfrac{x}{2}\right)^{2/3}$ - Limits: from $x=0$ to $x=2$ The arc-length of a curve is the length of the curve measured along it between two given points. For a smooth curve $y=f(x)$ from $x=a$ to $x=b$: $$L = \inta^b \sqrt{1+\left(\frac{...
Full solved answer →Area between curves
Define area between two curves. Find area of the region enclosed by the parabola $y = 2 - x^2$ and the line $y = -x$. Define volume integral. Find the volume of solid generated by revolving the region bounded by the curve $y^2 = x$ and the line $y = 1$, $x = 4$ about the line $y = 1$. [1+3+6]
The area between two curves is the region bounded between two curves in a plane over a common interval. If $y = f(x)$ and $y = g(x)$ with $f(x) \ge g(x)$ on $[a,b]$, then: $$A = \inta^b [f(x) - g(x)] \, dx$$ --- Step 1: Intersection points $$2 - x^2 = -x \i...
Full solved answer →Find the area between the curves $y = x^2 - 2$ and $y = 2$. [5]
- Curve 1 (parabola): $y = x^2 - 2$ - Curve 2 (horizontal line): $y = 2$ Step 1: Find intersection points Set the two expressions for $y$ equal: $$x^2 - 2 = 2$$ $$x^2 = 4$$ $$x = \pm 2$$ Intersection points at $x = -2$ and $x = 2$. Step 2: Determine the upp...
Full solved answer →Surface area of revolution
Find the area of the surface generated by revolving the curve $y = 2\sqrt{x}$, $1 \leq x \leq 2$, about the x-axis. [5]
- Curve: $y = 2\sqrt{x}$ - Interval: $1 \le x \le 2$ - Axis of revolution: x-axis $$S = 2\pi \inta^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ $$y = 2x^{1/2} \implies \frac{dy}{dx} = x^{-1/2} = \frac{1}{\sqrt{x}}$$ $$\left(\frac{dy}{dx}\right)^2 = \...
Full solved answer →Make Unit 7 stick
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