Mathematics I · Unit 5 · 5 hrs
Antiderivatives
Exam-focused notes for Antiderivatives (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 7 solved past questions from this unit.
What this unit covers
- Review of antiderivatives
- Rectilinear motion
- Indefinite integrals and Net change
- Definite integral
- The Fundamental theorem of calculus
- Improper integrals
Definite integral
Evaluate $\int_0^3 \sqrt{1 + x^2} \cdot x^3 , dx$ [5]
- Integrand: $\sqrt{1+x^2}\cdot x^3$ - Limits: from $x=0$ to $x=3$ Let $u = 1+x^2$, so $du = 2x\,dx$, and $x^2 = u-1$. Write $x^3\,dx = x^2\cdot x\,dx = (u-1)\cdot\dfrac{du}{2}$. Limits: $x$ $u=1+x^2$ ---------------- $0$ $1$ $3$ $10$ $$\int0^3 \sqrt{1+x^2}...
Full solved answer →Integrate $\int_0^1 x^2 \sqrt{x^3 + 1} dx$ [5]
Definite integral to evaluate: $$\int0^1 x^2\sqrt{x^3+1}\, dx$$ Lower limit $x=0$, upper limit $x=1$. Let $u = x^3 + 1$. $$\frac{du}{dx} = 3x^2 \implies du = 3x^2\, dx \implies x^2\, dx = \frac{du}{3}$$ - $x = 0 \Rightarrow u = 0^3 + 1 = 1$ - $x = 1 \Righta...
Full solved answer →Improper integrals
Show the integral coverages $\int_0^3 \frac{dx}{x-1}$. [5]
- Integral: $\displaystyle\int0^3 \frac{dx}{x-1}$ - Integrand: $f(x) = \dfrac{1}{x-1}$ - Limits: lower $= 0$, upper $= 3$ The denominator vanishes when $x - 1 = 0$, i.e. at $x = 1$. Since $1 \in (0,3)$, the integrand has an infinite discontinuity inside the...
Full solved answer →Show that the following integrals converge and diverge respectively.
$$\int_1^{\infty} \frac{1}{x^2} dx \text{ and } \int_1^{\infty} \frac{1}{x} dx$$
If $f(x,y) = \frac{xy}{x^2 + y^2}$, does $\lim_{(x,y) \to (0,0)} f(x,y)$ exist?
A particle moves in a straight line and has acceleration given by $a(t) = 6t^2 + t$. Its initial velocity is $4$ m/sec and its initial displacement is $s(0) = 5$ cm. Find its position function $s(t)$.
[2+3+5]
- Integrals: $\int1^{\infty} \frac{1}{x^2}\,dx$ and $\int1^{\infty} \frac{1}{x}\,dx$ - Function: $f(x,y) = \dfrac{xy}{x^2+y^2}$; limit as $(x,y)\to(0,0)$ - Acceleration: $a(t) = 6t^2 + t$ - Initial velocity: $v(0) = 4$ m/s - Initial displacement: $s(0) = 5$...
Full solved answer →Evaluate $\int_0^{\infty} x^3 \sqrt{1-x^4} dx$ [5]
- Integrand: $f(x) = x^3\sqrt{1-x^4}$ - Limits: from $0$ to $\infty$ For the square root $\sqrt{1-x^4}$ to be real, we need: $$1 - x^4 \geq 0 \implies x^4 \leq 1 \implies x \leq 1$$ On $[0,\infty)$ this means the integrand is real only for $0 \leq x \leq 1$...
Full solved answer →Determine whether the integral is convergent or divergent. $$\int_1^{\infty} \frac{1}{x} dx$$ [5]
- Integral: $\displaystyle \int1^{\infty} \frac{1}{x}\, dx$ - Lower limit: $1$, Upper limit: $\infty$ - Integrand: $\frac{1}{x}$ Since the upper limit is infinite, this is an improper integral of Type 1: $$\int1^{\infty} \frac{1}{x}\, dx = \lim{t \to \infty...
Full solved answer →Rectilinear motion
Using rectangles, estimate the area under the parabola $y = x^2$ from 0 to 1. A particle moves along a line so that its velocity v at time t is $v = t^2 + t + 6$. (i) find the displacement of the particle during the time period $1 \leq t \leq 4$. (ii) find the distance traveled during this time period. [5+5+0]
- Curve: $y = x^2$, interval $[0,1]$. - Velocity: $v(t) = t^2 + t + 6$, time interval $1 \le t \le 4$. --- (a) Area under $y=x^2$ from 0 to 1 (rectangle method) Divide $[0,1]$ into $n$ equal subintervals. - Width: $\Delta x = \dfrac{1}{n}$ - Right endpoints...
Full solved answer →Make Unit 5 stick
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