1. Since d/dx sinx = cosx, sinx=x-1/3!x^3+1/5!x^5-1/7!x^7…(-1)^k/(2k+1)!x^(2k+1) and we know cosx=1-1/2!x^2+1/4!x^4-…+(-1)^k/2k!x^2k. So the derivative of Maclaurin series of sinx equals to cosx
2. ln(x+1)=x-x^2/2+x^3/3…+(-1)^n/n+1 x^n+1. So we can ger derivative of ln(x+1) is 1-x+x^2-x^3+…+(-1)^nx^n. Since d/dxln(x+1)=1/x+1. We can find  the derivative of Maclaurin series for ln(x+1)equals to 1/x+1.
3. sin(x^2)=x^2-1/3!x^^+2/5!+(-1)^k4k+2/2k!x^4k+1, d/dx sin(x^2)=2xcos(x^2) We can find  the derivative of Maclaurin series for sin(x^2)equals to 2xcos(x^2).
4. for cosx=∑（-1）^n/2n!x^2n, we know d/dx cosx= -sinx. . We can find  the derivative of Maclaurin series for cosx equals to -sinx.

First question is ∫(0 to  square root pi)xsin(x^2)dx

Let u=x^2, then du=2xdx. The equation is ∫sin(u)/2 du=1/2(-cos(u))=-1/2cos(x^2). We can sub 0 to sqrt pi to equation, that is 1/2-(-1/2)=1

Using integeration by subsitution is a good way to solve the questiion.

We also can use integration by parts, dv=x v=1/2x^2 . u=sin(x^2) du=xconx. By formula udv=uv-∫vdu. By substitude0 to sqrt pi, we will get 1/2-(-1/2)=1.

Second question is ∫(0 to  square root pi)xcos(x^2)dx

Let u=x^2, then du=2xdx. The equation is ∫cos(u)/2 du=1/2(-sin(u))=-1/2sin(x^2). We can sub 0 to sqrt pi to equation, that is 0-0=1

We also can use integration by parts, dv=x v=1/2x^2 . u=cos(x^2) du=-xsinx. By formula udv=uv-∫vdu. By substitude0 to sqrt pi, we will get 0-0=1.

The third question  ∫(0 to  square root pi)x^3sin(x^2)dx. Let u=x^2, then du=2xdx. The equation is ∫usin(u)/2 du. Using integration by parts, u=u,u’=1,v’=sin(u) v=-cos(u). 1/2(-ucons(u)-(-∫cos(u)du))=1/2(sin(x^2)-x^2cos(x^2)).B y substitude0 to sqrt pi, we will get pi/2-0=pi/2.

The method is used by integration by parts and substitution.

Many knowledge is connected in math. In my opinion, integration is expressed by anti-derivative form. anti-derivative is the method to do integration. So some students may confused with these two concepts, since they might think they are basically same.

Anti-derivative is the inverse operation of the derivative, which in order to find what value results this result. This is the method via derivative to get the answer.

Integration is the opposite concept with derivative. The integration is to find the area under the certain interval or uncertain interval.

Some students may think they can just use anti-derivative to represent the integration. However anti-derivative does not have interval.

According to these two concepts, we can see that anti-derivative is the method to solve integration problem, since students need find the integration via anti-derivative which can represent the uncertain interval result for integration. So anti-derivative is the crucial way to calculate the  integration.

Math can change the world. The motivation of developing   integer is from the needs of application. In reality, sometimes people can eliminate some unknown variable, but with the development of technique. It makes more and more important to know the exact value. For same simple figures, such as rectangle, circle, triangle… We can use formula to find the area. However, if we need know some complex figures. We have to use integer.

Basically, integer is divide the figure into some parts. That we can refer these some parts as some small rectangles. We can easily find the area of the rectangle (width times length) instead of the complex figure. We just need to add these rectangles together. Then we can get the area of the complex figure.

This method is created by Riemann in 1854. He had a speech to illustrate his method to find the area of complex figure. His discover not only impact the geometry, but also form the fundamental of the Einstein’s relativity. We can literally say, Riemann change the world.

Lebesgue integration solves the problem that Riemann’s integration cannot solve. It moves the integer from 2-d to every dimensions.

In question 5, the  most cruical thought is to find the relationship between each block. We know if we want to find the sum of something, we use a pattern to express this relationship by writing the general formule.  By finding the general formule, we can use the thought of limit to find the extreme value, such as when the last term is infinity. We could use the formule or equation to find it. Also, we should know, in reality, many extreme situations are assumed by us. So, it is impossible to really find the exact value for the infinity term .( Which means we can find someting that still break the rule.)

a)For last term {an} of the sequence ,{an} less or equal to a constant, the sequence converged.If not,  the sequence diverged. The convergent sequence is like a road, which can arrive the destination(the final term), eventually, for example {0,0,0,0,0,0,0,0}. The divergent sequence is like you are traveling in the universe, you will not find the end(the final term approaches to infinity),for example,{1,-1,1,-1,1,-1,1,-1,1,-1}. Basically, the trent of convergent sequence will approach to a constant, the trent of divergent sequence approach to not approch to a constant(+-inifity or up and down).

b)For the convergent series, the sum of these term will approach to a constant.If not, the series will be divergent(+-inifity or up and down). So the idea is to calculate the sum of the series is close to a constant or not. So the convergent series is like a hungry breasts to approach to their preies,but they will not really catch the preies, for example , the sum of {1\n^2},n =[1,inifity). The divergent series is like the preies, run away from the breasts’ mouth,for example, the sum of{1\n},n=[1,inifity).

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