Foundation Mathematics
1017SCG
Week 11
Topics
- Introduction to Integration
- Tables of Integrals
- Definite Integrals (Area Under Curve)
Introduction to Integration
Consider the function:
$\;y= 3x^2-4x+5$
Recall how we computed the derivative
$\ds \frac{dy}{dx}$ $= 6x-4$
Introduction to Integration
$\;y= 3x^2-4x+5$
Question: If we know the derivative
$\ds \frac{dy}{dx}$ $= 6x-4$
is it possible to recover back $y$?
👉 $\;y = 3x^2-4x+C,$ with $C$ a constant.
Introduction to Integration
Thus, if $\,\ds \frac{dy}{dx} = 6x-4,\,$ we have $\,\ds y = 3x^2-4x + C$
$\ds \int$ $ 6x - 4$ $dx$ $ = 3x^2-4x + C$
This process is called:
Anti-differentiation = Integration
Introduction to Integration
A process developed in the late 1700s
Anti-differentiation = Integration
Isaac Newton |
Gottfried Wilhelm Leibniz |
Big question
Can we compute the integral
of any given function $f(x)$?
The answer is "Yes",
as long as the function $f(x)$ is
continuous.
Warning!
Remark about computing anti-derivatives❗️
Although the theory says the integral
$ \ds \int f(x)\, dx = F(x) + C $
exists for $f(x)$ continuous functions,
we cannot always find a formula for $F(x)$
Remark about computing anti-derivatives❗️
Although the theory says the integral
$
\ds \int f(x)\, dx = F(x) + C
$
exists for $f(x)$ continuous functions,
we cannot always find a formula for $F(x)$
Here are some examples:
$\ds\int e^{-x^2}\, dx,\;$ $\ds\int\sin\left(x^2\right)\, dx,\;$ $\ds\int\cos\left(x^2\right)\, dx$
These are known as non-elementary integrals
Good news! This topic is outside the scope of this course.
Here we consider only basic, elementary functions! 😃
Table of integrals
| $f(x)$ | $\ds \int f(x)\,dx$ |
| $x^n$ | $\dfrac{x^{n+1}}{n+1}+C$ |
| $ax^n$ | $\dfrac{ax^{n+1}}{n+1}+C$ |
| $\sin(x)$ | $-\cos(x)+C$ |
| $\cos(x)$ | $\sin(x)+C$ |
| $e^x$ | $e^x+C$ |
| $\dfrac{1}{x}$ | $\ln |x|+C$ |
Examples: $x^n$
• $\ds \int x^2 \,dx$ $= \dfrac{x^{2+1}}{2+1}+C$ $= \dfrac{x^{3}}{3}+C$
• $\ds \int x^4 \,dx$ $= \dfrac{x^{4+1}}{4+1}+C$ $= \dfrac{x^{5}}{5}+C$
Examples: $a x^n$
• $\ds \int 2x^4 \,dx$ $= \dfrac{2x^{4+1}}{4+1}+C$ $= \dfrac{2x^{5}}{5}+C$
• $\ds \int 6x \,dx$ $= \dfrac{6x^{1+1}}{1+1}+C$ $= \dfrac{6x^{2}}{2}+C$
• $\ds \int 2x^2-4x+5 \,dx$ $= \dfrac{2x^{3}}{3}$ $-\,\dfrac{4x^{2}}{2}$ $+\,5x$ $+\,C$
Examples: $a\sin(x),\,b \cos(x)$
•$\ds \int 2\cos(x) \,dx$ $=2\sin(x)+C$
•$\ds \int 3\sin(x) \,dx$ $=3(-\cos(x))+C$ $=-3\cos(x)+C$
•$\ds \int -7\cos(x) + 3\sin(x) \,dx$ $= -7\sin(x)$ $-3\,\cos(x)+C$
Examples: $ae^x$ and $a/x$
• $\ds \int 2e^x \,dx$ $=2e^x+C$
• $\ds \int \frac{4}{x} \,dx$ $=4\ln|x|+C$
• $\ds \int -7e^x + \frac{3}{x} \,dx$ $= -7e^x$ $+\,3\ln|x|+C$
One more example
$\ds \int x^2 + 2\sin(x)+ 3\cos(x)+ 5e^x +\frac{7}{x} \,dx$
$\ds = \frac{x^3}{3}$ $+\,2(-\cos(x))$ $+\,3\sin(x)$ $+\,5e^x$ $+\,7\ln|x|$ $+\,C$
$\ds = \frac{x^3}{3}$ $-\,2\cos(x)$ $+\,3\sin(x)$ $+\,5e^x$ $+\,7\ln|x|$ $+\,C$
Definite Integrals
Consider the continuous function $f(x)$ on $[a,b]$ and $F(x)$ is function such that $F'(x)=f(x)$ for every $x\in[a,b],$ then
• We call $F(x)$ an anti-derivative.
• This fact is known as The Fundamental Theorem of Calculus.
Definite Integrals: Example 1
$\ds \int_1^3 x^2 \, dx$ $\ds = \left[\dfrac{x^3}{3}+C\right]_1^3$ $\qquad \quad \quad\; F(x) = \dfrac{x^3}{3}+C\;$ 👈
$\qquad \qquad \ds = \underbrace{\left[\dfrac{(3)^3}{3}+C\right]}_{F(3)} - \underbrace{\left[\dfrac{(1)^3}{3}+C\right]}_{F(1)}$
$\qquad \qquad \ds = \dfrac{27}{3}+C $ $ -\, \dfrac{1}{3}-C$
$\qquad \qquad \ds = \dfrac{26}{3}$
Definite Integrals: Example 1
$\ds \int_1^3 x^2 \, dx$ $\ds = \left[\dfrac{x^3}{3}\right]_1^3$ $\qquad \quad \quad\; F(x) = \dfrac{x^3}{3}+C\;$ 👈
$\qquad \qquad \ds = \underbrace{\left[\dfrac{(3)^3}{3}\right]}_{F(3)} - \underbrace{\left[\dfrac{(1)^3}{3}\right]}_{F(1)}$
$\qquad \qquad \ds = \dfrac{27}{3} $ $ -\, \dfrac{1}{3}$
$\qquad \qquad \ds = \dfrac{26}{3}$
👉 Note we don't need to write the constant $C$.
Definite Integrals: Example 1
|
$\ds \int_1^3 x^2 \, dx= \dfrac{26}{3}$ |
$f(x)=x^2$ |
Definite Integrals: Example 2
|
$\ds \int_{-3}^3 2x \, dx $ $=\ds \left[x^2+C\right]_{-3}^3$ $\quad =\ds \left[(3)^2+C\right]-\left[(-3)^2+C\right]$ $\quad =\ds \left[9+C\right]-\left[9+C\right]$ $\quad =\ds 9+C-9-C$ $\quad =\ds 0$ |
$f(x)=2x$ |
Definite Integrals: Example 3
$\ds \int_{0}^{\frac{3\pi}{2}} 2\cos(x) \, dx $ $=\ds \bigg[2\sin(x)\bigg]_0^{\frac{3\pi}{2}}$ $ =\ds 2\sin\left(\frac{3\pi}{2}\right)-2\sin\left(0\right)$
$\qquad \qquad\qquad=\ds 2(-1)-2(0)$ $ =\ds -2$
Definite Integrals: Example 3
If we want to find the total area bounded by the curve $y=2\cos(x)$ and between $x=0$ and $x=\dfrac{3\pi}{2},$ then we need to compute two integrals:
$\ds \int_{0}^{\frac{\pi}{2}} 2\cos(x) \, dx$ $\ds=2\;\;$ and $\;\; \ds \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2\cos(x) \, dx$ $=-4$
Definite Integrals: Example 3
Therefore, the total area (TA) bounded by the curve $y=2\cos(x)$
and between $\,x=0\,$ and $\,x=\dfrac{3\pi}{2}\,$
is:
TA = $\ds\left|\int_{0}^{\frac{\pi}{2}} 2\cos(x) \, dx\right|+\left|\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2\cos(x) \, dx\right| $ $=|2|+ |-4|$ $=6$
Final remarks
From Calculus we have
Integration and Differentiation have been key
to the development of modern science and technology.
Final remarks
The invention of Calculus is often attributed to
|
Isaac Newton |
Gottfried Wilhelm Leibniz |
However, this is an oversimplification, as Calculus is the result of a long evolution in which both played a decisive role.
Final remarks
Newton and Leibniz were highly criticised! 🤬
Let $y = x^2.\,$ Using Newton & Leibniz's method to compute the derivative we consider $h$ to be an infinitesimal increment:
$\ds \dfrac{f(x+h)-f(x)}{h} $ $\ds = \dfrac{\left(x^2 + 2xh + h^2\right) -\left(x^2\right)}{h} $ $ \ds = \ds \dfrac{ 2xh + h^2}{h}$
$\ds = \dfrac{ \left(2x + h\right)h}{h}$ $\ds =2x+h .\;$ Therefore $\;\ds \frac{dy}{dx} = 2x$
George Berkeley (1685-1753) argued that it was logically inconsistent to introduce an infinitesimal increment $h$ in order to perform calculations, and then to let $h$ vanish at the end of the process.
Final remarks
Fortunately, the issue was solved a few years later with
the introduction of the formal definition of limit.
Let $y = x^2.\,$ Using limits we have
$\ds \lim_{h\ra 0} $ $ \dfrac{f(x+h)-f(x)}{h}$ $\ds = $ $\ds \lim_{h\ra 0} $ $\ds\dfrac{\left(x^2 + 2xh + h^2\right) -\left(x^2\right)}{h} $
$\qquad \ds = $ $\ds \lim_{h\ra 0} $ $\ds \dfrac{ 2xh + h^2}{h}$ $\ds = $ $\ds \lim_{h\ra 0} $ $\ds \dfrac{ \left(2x + h\right)h}{h}$ $\ds = $ $\ds \lim_{h\ra 0} $ $\ds\left(\right. $ $\ds 2x+h$ $\ds\left.\right) $ $\ds=2x$
Now we can properly say that $\ds\frac{dy}{dx}=2x.$
Final remarks
|
Isaac Newton |
Gottfried Wilhelm Leibniz |
The Calculus, next to the Euclidian geometry, is the greatest creation in all mathematics.
Moris Kline, 1972.
References
- Boyer, C. B. (1949). The History of the Calculus and its Conceptual Development. Dover Publications, Inc. New York.
- Edwards, C. H. (1979). The Historical Development of the Calculus. New York Springer-Verlag.
- Kline, M. (1972). Mathematical Thought from Ancient to Modern Times. New York. Oxford University Press.
- Whiteside, D. T. (1960). Patterns of mathematical thought in the later seventeenth century. Archive for History of Exact Sciences. 1, 179-388.