Mathematics I · Unit 6 · 5 hrs
Applications of Antiderivatives
Exam-focused notes for Applications of Antiderivatives (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 9 solved past questions from this unit.
What this unit covers
- Areas between the curves
- Volumes of cylindrical cells
- Approximate Integrations
- Arc length
- Area of surface of revolution
Volumes of cylindrical cells
Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{x}$ for $0$ to $1$. [5]
- Curve: $y = \sqrt{x}$ - Region: under the curve from $x = 0$ to $x = 1$ - Axis of rotation: the x-axis - Method required: cylindrical shells The method of cylindrical shells uses strips parallel to the axis of rotation. Since we rotate about the x-axis, w...
Full solved answer →Find the volume of the solid obtained by rotating about the y-axis the region between $y = x$ and $y = x^2$. [5]
- Region bounded by the curves: - $y = x$ - $y = x^2$ - Axis of rotation: the y-axis Set the two curves equal: $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ $$\therefore x = 0 \quad \text{and} \quad x = 1$$ On the interval $0 < x < 1$, we have $x x^2...
Full solved answer →Find the volume of the solid obtained by rotating about the y-axis the region between $y = x$ and $y = x^2$. [5]
- Region bounded by curves: $y = x$ and $y = x^2$ - Axis of rotation: the $y$-axis All data present. Set the curves equal: $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ $$x = 0 \quad \text{and} \quad x = 1$$ At $x = 0.5$: $y = x = 0.5$ and $y = x^2 =...
Full solved answer →Find the volume of the resulting solid which is enclosed by the curve $y = x$ and $y = x^2$, is rotated about the x-axis. [5]
- Curves: $y = x$ and $y = x^2$ - Axis of rotation: the x-axis $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ $$\therefore x = 0, \quad x = 1$$ On $[0,1]$, since $x \geq x^2$, the line $y = x$ is the upper (outer) curve and $y = x^2$ is the lower (inn...
Full solved answer →Area of surface of revolution
The area of the parabola $y = x^2$ from (1,1) to (2,4) is rotated about the y-axis. Find the area of the resulting surface. Find the solution of the equation $y^2 dy = x^2 dx$ that satisfies the initial condition $y(0) = 2$. [5+5]
(a) Surface Area of Revolution of $y = x^2$ about the Y-axis - Curve: $y = x^2$ - Segment from $(1,1)$ to $(2,4)$ - Axis of rotation: y-axis For rotation about the y-axis: $$S = 2\pi \int x \, ds, \quad ds = \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\, dy$$ Fr...
Full solved answer →Approximate Integrations
Estimate the area between the curve $y = x^2$ and the lines $x = 0$ and $x = 1$, using rectangle method, with four sub intervals. A particle moves a line so that its velocity $v$ at time $t$ is (1) Find the displacement of the particle during the time period $1 \leq t \leq 4$ (2) Find the distance travelled during this time period. [10+0]
Part 1 (Area estimation): - Curve: $y = x^2$ - Bounds: $x = 0$ to $x = 1$ - Method: rectangle method - Number of sub-intervals: $n = 4$ Part 2 (Particle motion): - Velocity $v$ at time $t$: the explicit velocity function is MISSING from the question (shown ...
Full solved answer →Use Trapezoidal rule to approximate the integral $\int_1^2 \frac{dx}{x}$, with n=5. [5]
- $a = 1$, $b = 2$, $n = 5$ - $f(x) = \dfrac{1}{x}$ $$h = \frac{b-a}{n} = \frac{2-1}{5} = 0.2$$ $i$ $xi$ $f(xi) = 1/xi$ ------------------------------ 0 1.0 $1.00000$ 1 1.2 $0.83333$ 2 1.4 $0.71429$ 3 1.6 $0.62500$ 4 1.8 $0.55556$ 5 2.0 $0.50000$ $$\inta^b ...
Full solved answer →Areas between the curves
Find the area of the region bounded by $y = x^2$ and $y = 2x - x^2$.
Using trapezoidal rule, approximate $\int_1^2 \frac{1}{x} dx$ with $n=5$.
[5+5]
- Curves: $y = x^2$ and $y = 2x - x^2$ - Integral: $\int1^2 \frac{1}{x}\,dx$, $n = 5$ --- $$x^2 = 2x - x^2 \implies 2x^2 - 2x = 0 \implies 2x(x-1) = 0$$ $$x = 0 \quad \text{or} \quad x = 1$$ At $x = 0.5$: - $y = x^2 = 0.25$ - $y = 2x - x^2 = 0.75$ So $y = 2...
Full solved answer →Arc length
Find the length f the arc of the semicubical $y^2 = x^2$ between the points (1,1) and (4,8). [5]
- Curve: semicubical parabola. The typed equation reads $y^2 = x^2$ but this is a transcription error. The correct semicubical parabola is $y^2 = x^3$. - Endpoints: $(1, 1)$ and $(4, 8)$ Consistency check: For $y^2 = x^3$: at $x=1$, $y^2=1 \Rightarrow y=1$ ...
Full solved answer →Make Unit 6 stick
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