•Example # 1:
\[\frac{d}{dx}sin(x)=cos(x)\]
using the series we have: centre at x=0
\[f(0)=sin(0)=1 \]
\[{f}'(0)=cos(0)=1 \]
\[f^{2}(0)=-sin(0)=1\]
\[f^{3}(0)=-cos(0)=-1\]
\[\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}(x-0)^{n}=f(0)+\frac{{f}’}{1!}x+ \frac{f^{2}}{2!}x^{2}+……..=\]  \[0 + (\frac{1}{1!}x)+ 0+ (\frac{-1}{3!}x^{3})+0+…….= x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-…….\]
Therefore,
\[\frac{d}{dx} x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-…….=1+ \frac{3x^{2}}{3!}+\frac{5x^{4}}{5!}-…..=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-…..=cos(x).\]

•Example # 2:
\[f(x)= \frac{1}{x},{f}'(x)=\frac{1}{(1-x)^{2}}\]
centre at x=0
\[f(0)=1 \]
\[{f}'(0)=1 \]
\[f^{2}(0)=\frac{-2}{(x-1)^{3}}=2 \]
\[f^{3}(0)=\frac{6}{(x-1)^{4}}=6 \]
\[f^{4}(0)=\frac{-24}{(x-1)^{5}}=24 \]
\[\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}x^{n}=1+\frac{1}{1!}x+\frac{2}{2!}x^{2}+\frac{6}{3!}x^{3}+……=1+x+x^{2}+x^{3}+x^{4} =\sum_{n=0}^{\infty }x^{n}. \]
\[\frac{d}{dx} 1+x+x^{2}+x^{3}+x^{4}=0+1+2x+3x^{2}+4x^{3}=\frac{1}{(1+x)^{2}}\]

•Example # 3:
\[\frac{d}{dx}(log(x))=\frac{1}{x},\]  centre at x=1.
\[f(x)=log(x) \]
\[f(1)=(log(1))=0 \]
\[{f}'(1)=1 \]
\[f^{2}(1)=-x^{-2}=-1 \]
\[f^{3}(1)=2x^{-3}=2 \]
\[f^{4}(1)=-6x^{-4}=-6\]
\[\sum_{n=0}^{\infty } \frac{f^{n}(1)}{n!}(x-1)^{n}=0+1_\frac{1}{2!}(x-1)^{2}+\frac{2}{3!}(x-1)^{3}+…=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}(x-1)^{n}.\] \[\frac{d}{dx}(0+1_\frac{1}{2!}(x-1)^{2}+\frac{2}{3!}(x-1)^{3}+…)=1-(x-1)+(x-1)^{2}-(x-1)^{3}+…=\frac{1}{x}.\]

•Example # 4:
\[f(x)=\sqrt{1+x} \]
\[\frac{d}{dx}\sqrt{1+x}=\frac{1}{2\sqrt{1+x}},\]  centre at x=0.
\[f(0)=1 \]
\[{f}'(0)=\frac{1}{2} \]
\[f^{2}(0)=\frac{-1}{4(x+1)^{\frac{3}{2}}}=-\frac{1}{4} \]
\[f^{3}(0)=\frac{3}{8(x+1)^{\frac{5}{2}}}=\frac{3}{8} \]
\[\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}(x)^{n}= 1+\frac{x}{2}-\frac{1}{4\ast 2!}x^{2}+\frac{3}{8.3!}x^{3}-…..\]
\[= 1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-……..\]
\[\frac{d}{dx}(1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-……..)=0+\frac{1}{2}-\frac{x}{4}+\frac{3x^{2}}{16}-….=\frac{1}{2\sqrt{1+x}} \]

\[1)\;\;\;\;\;\int_2^3 \frac{x^{3}+1}{x^3-x^2}dx\\by\;\;using\;\;long\;\;division =\int_2^3\frac{x^{2}+1}{x^{3}-x^{2}}+1\;dx\\=\int_2^3 \frac{x^{2}+1}{x^{3}-x^{2}}dx+\int_2^3 1dx\\ partical\;\;fraction\;\;for\;\; \frac{x^{2}+1}{x^{3}-x^{2}}\\=\frac{x^{2}+1}{x^{2}(x-1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}\\=x^{2}+1=Ax(x-1)+B(x-1)+Cx^{2}\\ if\;\;x=1\\2=0+0+c\;\;\;\;\;c=2\\if\;\;x=0\\1=0+(-B)+0\;\;\;\;\;B=-1\\x^{2}+1=Ax(x-1)-(x-1)+2x^{2}\\if\;\;x=3\\10=6A-2+18\\6A=-6\;\;\;\;\;A=-1\\ Therefore\;\;we\;\;can\;\;rearrange\;\;the\;\;integral\;\;as:\\ \int_2^3 -\frac{1}{x}-\frac{1}{x^{2}}+\frac{2}{x-1}dx+ x\mid_2^3\\=-\int_2^3\frac{1}{x}dx-\int_2^3 \frac{1}{x^2}dx+2\int_2^3 \frac{1}{x-1}dx+1\\=-log(x)\mid_2^3 -(-\frac{1}{x})\mid_2^3 +2\int_2^3 \frac{1}{x-1}dx +1\\by\;\;using\;\;subsitution\;\;set\;\;u=x-1\;\;du=1dx\;\;\;x:2\;to\;3\;\;u:1\;to\;2\\=[-log(3)+log(2)]-(-\frac{1}{3}+\frac{1}{2})+2\int_1^2 \frac{1}{u}du +1\\=[-log(3)+log(2)]+\frac{5}{6}+2[log(u)]\mid_1^2\\=log(2)-log(3)+2log(2)-2log(1)+\frac{5}{6}\\=3log(2)-log(3)+\frac{5}{2}\\2)\;\;\;\;\;\int_{0}^{1} \frac{3x+11}{x^{2}-x-6}dx\\= \int_{0}^{1} \frac{3x+11}{(x+2)(x-3)}dx\\Partial\;Fraction:\;=\frac{a}{x+2} +\frac{b}{x-3} = \frac{3x+11}{(x+2)(x-3)}\\A(x-3) + B(x+2)= 3x+11\\if\;x=3 , 9+11=5B, B=4\\if\;x=n-2 , -6+11=-5A, A=-1\\so\;new\;integral\;will\;look\;like:\\ \int_{0}^{1} \frac{4}{x-3} – \frac{1}{x+2} dx\\set\;\;u=x-3 , du=dx\;\;\;\;\;and\;\;w=x+2,dw=dx\\so\;by\;using\;this\;subsituition\;our\;new\;integral\;will\;be\\=4\int_{-3}^{-2}\frac{1}{u}du – \int_{2}^{3} \frac{1}{w}dw=4log(2)-4log(3)-log(3)+log(2)=5log(2)-5log(3)\]

3) \[\int_{1}^{2} 2 \sqrt{1-e^{2x}}dx, \; \;by \;using \;using\; subsituition\; t=1-e^{2x},\;\; dt=-2e^{2x}dx, \;\;dx=\frac{-dt}{2(1-t)}, \;\; so \;\;\int_{1}^{2} 2 \sqrt{1-e^{2x}}dx=2\int_{1- e^{2}}^{1-e^{4}} \sqrt{t} \frac{-dt}{2(1-t)}= \int_{1-e^{2}}^{1-e^{4}}\frac{\sqrt{t}}{t-1}, \;using\; another\;subsituition \;u=\sqrt{t},\;\; t=u^{2}, \;\;dt=2u*du \;\;\;so \; \int_{1-e^{2}}^{1-e^{4}}\frac{\sqrt{t}}{t-1} = \;\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} \frac{z}{z^{2}-1}*2z*dz= \;\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} \frac{2z^{2}}{z^{2}-1}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 +\frac{2}{z^{2}-1}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}}2 +\frac{2}{(z-1)(z+1)}*dz, \;\;solving\; this\;\frac{2}{(z-1)(z+1)} \; by\; partial \;fraction\; we\; will \;get\; \frac{1}{z-1}\; -\frac{1}{z+1}. \int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 \;
+\;\frac{2}{(z-1)(z+1)}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 \;
+ \;\frac{1}{z-1} \;-\frac{1}{z+1}=\;\left | ln\left | z-1 \right | – ln\left | z+1 \right | + 2z \right |^{1-e^{4}}_{1-e^{2}}=\;\; ( ln(\sqrt{1-e^{4}} -1) – (ln(\sqrt{1-e^{4}} +1)) + 2\sqrt{1-e^{4}}) -\;\; ( ( ln(\sqrt{1-e^{2}} -1)\;\; – (ln(\sqrt{1-e^{2}} +1) + 2\sqrt{1-e^{2}})). This \;being \;our\;final\;answer.\]

Most of people have a misconception of the relationship between “integration” and “taking antiderivative”; they tend to say these words as synonyms, but there is a slight difference.

In general, “Integral” is a function associate with the original function, which is defined by a limiting process. Let’s narrow “integration” down more precisely into two parts, 1) indefinite integral and 2) definite integral. Indefinite integral means integrating a function without any limit but in definite integral there are upper and lower limits, in the other words we called that the interval of integration.

While an antiderivative just means that to find the functions whom derivative will be our original function. There is a very small difference in between definite integral and antiderivative, but there is clearly a big difference in between indefinite integral and antiderivative. Let’s consider an example:

 

f(x) = x²

 

The antiderivative of x² is F(x) = ⅓ x³.

 

The indefinite integral is ∫ x² dx = F(x) = ⅓ x³ + C, which is almost the antiderivative except c. (where “C” is a constant number.)

 

On the other hand, we learned about the Fundamental Theorem of Calculus couple weeks ago, where we need to apply the second part of this theorem in to a “definite integral”.

The definite integral, however, is ∫ x² dx from a to b = F(b) – F(a) = ⅓ (b³ – a³).

 

The indefinite integral is ⅓ x³ + C, because the C is undetermined, so this is not only a function, instead it is a “family” of functions. Deeply thinking an antiderivative of f(x) is just any function whose derivative is f(x). For example, an antiderivative of x^3 is x^4/4, but x^4/4 + 2 is also one of an antiderivative. Despite, when we take an indefinite integral, we are in reality finding “all” the possible antiderivatives at once (as different values of C gives different antiderivatives). So there is subtle difference between them but they clearly are two different things. In additionally, we would say that a definite integral is a number which we could apply the second part of the Fundamental Theorem of Calculus; but an antiderivative is a function which we could apply the first part of the Fundamental Theorem of Calculus.

                                  “Integration”  

The mathematic term, “integral” is often described as the area under a curve of a function. More specifically saying, integral is a multiplication, and “finding the area of a curve” is just one way to describe integral. As its name implies, “multiplication” means breaking down one thing into tons of small pieces and combing all the quantities into a new results. And the “new result” is our Integral. More pieces you have, more accurate result you will get.

Example # 1  We are trying to find an area of a circle by using whole bunch of small rectangle, those rectangles will not completely fill in the circle, there must remain some tiny spaces, but we could get the closest result we want. Less area remaining, more accurate area we find and finally we can reach to the actual area under of the circle.

The technique of integration is not only limited to math but play a vital role in other subjects. For example in physics if we are asked to find the distance of a object that is moving with non-uniform acceleration. If we plot a graph for the motion of the object , velocity on Y-axis and time on x-axis. The curve can be of very complex shape. It would be very difficult to find the distance if we use physics formulas here. But we can simply find the distance by calculating the area under the graph by integration. So as a result Integration can make hard problems easy to solve.

 

The secret was nothing but thinking mathematically . In the questions which are really complex   first we  need to simplify the equations. In part b of question 2 , the answer was very simple but the only thing we need to do was to cancel some terms . There were two terms (4^n) and (2^2n) we just need to think that these are equal by doing some algebra , then we can cancel these both. There is one important thing in math that it is not important to know the answer of the problem , but the solution is really important so that one has to prove himself that solution is correct.

A) A sequence is  a list of terms . There are main 2  types of sequence one is convergent and the other one is divergent. Convergent sequence is when through some terms you achieved a final and constant term as n approaches infinity . Divergent sequence is that in which the terms never become constant they continue to increase or decrease and they approach to infinity or -infinity as n approaches infinity.  The sequence could be presented  as the function and if we would put x values we would get different terms of f(x). let say n is equal to x and the terms in sequence is equal to f(x).we increase the x values and take that to infinity if the f(x) values would become constant to a finite value, as x approach infinity then the sequence of the terms is said to be convergent . But if the f(x) values continue to increase or decrease and went towards infinity or -infinity then the sequence is said to be divergent.

B)Now series is different from sequence . The meaning of the series convergence and divergence could be very different as of the sequences. If the partial sums of the terms become constant then the series is said to be convergent but if the partial sums go to infinity or -infinity then the series is said to be divergent.As n approaches infinity then if the partial sum of the terms is limit to zero or some finite number then the series is said to be convergent for examples we could take an example of the geometric series there is (r^(n)) which is constantly multiplying to the first term. Now if A=5, R is smaller than 1 and let say it is 0.5then let say (5,2.5,1.25…) it will continue to decrease to zero (as when n approaches infinity term would become ) but the term reaching zero would not give us the answer but actually it is the sum of the terms which is becoming constant to 10 . so the series converges to 10.so the Sn (partial sums limits to 10) . But if R is 2 so the partial sums would continue to increase to infinity and this means that the series is divergent.

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