How to prove that a function is unbounded on an interval Prove that y = 1 / X sin1 / X is unbounded on (0,1]

How to prove that a function is unbounded on an interval Prove that y = 1 / X sin1 / X is unbounded on (0,1]

That is to say, for any large M, there is always a point X in the interval such that f (x) > M

Definite integral of discontinuous function
There is a question f (x) = √ 1-x & # 178; (- 1 "X" 1) f (x) = e to the x power (x > 1), then what is the definite integral of F (x) from - 1 to 2? Is it wrong to give the answer to π - 2? Shouldn't it be π - 2 + E & # 178; - e divided into - 1 to 1 and 1 to 2

Well, I think the answer to the question is wrong
The half circle area π / 2, the integral of the other section should be e ^ x | = e ^ 2-e,
My answer is the same as yours

What is the condition of upper function minus lower function in the application of geometric meaning of definite integral?

The most basic thing is that in the book, if you are continuous in a closed interval, you use the Niulai formula, which means that if you are continuous in a closed interval, there must be an original function. In essence, it connects the indefinite integral with the definite integral. In fact, the open interval can also be used under certain conditions. I will not elaborate on this

If the function f (x) is defined on [a, b] and | f (x) | is integrable on [a, b], why does the definite integral of F (x) not necessarily exist on [a, b]?
This is the first (4) fill in the blanks of exercise 5 on page 269 of higher mathematics of Tongji University
But I don't understand why

If f (x) is continuous on [a, b], then f (x) is integrable on [A.B]
Let me give you a counterexample,
Let f (x) = 1 (x ≥ 0), - 1 (x < 0) (a piecewise function form)
In this case, f (x) is not a continuous function, but | f (x) | = 1 is a continuous function
So f (x) is not necessarily integrable

When x → 0, a: 1-cosx B: 2x + TGX C: √ x + SiNx D: √ (1 + x) - √ (1-x)

C and D are both

The number of female students in class 6 (2) accounts for 4% of the total number of students in the class. What percentage is the number of female students less than that of male students?

Boys = 1-4 / 9 = 5 / 9
How many percent less girls than boys?
（5/9-4/9）÷5/9×100%=20%

32 times 78% minus 1.32 times 33% by a simple method

1.32*(78-33)/100=0.594

It is known that each item of the arithmetic sequence {an} is positive, A1 = 3, the sum of the first n terms is Sn, {BN} is an arithmetic sequence, the common ratio is q = 2, and a2b2 = 20, a3b3 = 56 (1) find an and BN (2) find the first n terms and TN of the sequence {anbn}
I hope the process is more detailed and urgent!

(1) an=a1+(n-1)d=3+(n-1)d
bn=b1*q^(n-1)=b1*2^(n-1)
a2b2=(3+d)*2b1=20
a3b3=(3+2d)*4b1=56
d=2 b1=2
an=2n+1
bn=2^n
(2) anbn=(2n+1)*2^n
Tn=3*2+5*2^2+7*2^3+...+(2n+1)*2^n
2Tn=3*2^2+5*2^3+...+(2n-1)*2^n+(2n+1)*2^(n+1)
Tn-2Tn=3*2+2(2^2+2^3+...+2^n)-(2n+1)*2^(n+1)
-Tn=6+8[1-2^(n-1)]/(1-2)-(2n+1)*2^(n+1)
Tn=(2n+1)*2^(n+1)-6+8-2*2^(n+1)
=(2n-1)*2^(n+1)+2

A few simple formulas: 1, 10-2.4 △ 0.3x0.62, 12.5-7 / 9 + 7.5-11 / 9 3, 5 / 11x [(2 / 5 + 1 / 3) △ 5 / 6]

1、10-2.4÷0.3x0.6
=10-8x0.6
=10-4.8
=5.2
2、12.5-7/9+7.5-11/9
=12.5+7.5-(7/9+11/9)
=20-2
=18
3、5/11x[(2/5+1/3)÷5/6]
=5/11x(2/5x6/5+1/3x6/5)
=5/11x(12/25+2/5)
=12/55+10/55
=22/55
=2/5

2-2-4-8-14-26-48-88 -?
What is a question mark?

The first three add 26 + 48 + 88 = 162