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Let the definite integral f(x)dx in the equation be t
Since f(x) is continuous, the integrals can be determined on both sides of the equation at the same time, and if the interval is [0,1], then the original function of t = arctan(1) -arctan(0) +t 41 (1+x 2) is arctan(x), and the original function of x 3 is x 4 4).
i.e. t = 3
Substituting the equation is the formula f(x).
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Convention: [a,b] denotes the definite integral on [a,b].
Let [0,1]f(x)dx=a (a is a constant) f(x)=1 (1+x)+ax
A= [0,1]f(x)dx
0,1](1 (1+x )+ax )dx [0,1](1 (1+x ))dx+ [0,1](ax )dx[0,1](1 (1+x ))dx+a [0,1]x dx and [0,1](1 (1+x ))dx=arctanx|[0,1]=π/4
0,1]x³dx=(1/4)x^4|[0,1]=1 4 gives a = ( 4) + (1 4).
The solution gives a= 3
So f(x)=1 (1+x )+ 3)x hope it helps!
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Summary. The definite integral is mainly to find the original function of the integral first, and then substitute the upper and lower bounds for calculation, and use the Newton-Leibniz formula to obtain the result.
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The dx (1+ x)=2 x-2ln(1+ x)+c definite integral is mainly to find the original function of the integral first, and then substitute the upper and lower bounds for calculation, and the result is obtained by using the Newton-Leibniz formula.
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It is the basic operation of the group cavity stove of the fixed circle and the integral, such as the following figure.
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You can use a formula, from the original definition of the Riemann integral (on the right is the limit of some kind of darboux sum):
a,b] f(x) dx=lim∑ f(a+(b-a)k/n)*(b-a)/n
The first [0,1] 1 (1+x 2) dx:
where a=0, b=1, so the integral value.
lim∑ f(k/n)/n
lim∑ (1/[1+(k/n)^2])/n
lim∑ n/(k^2+n^2)
lim n/(1^2+n^2)+n/(2^2+n^2)+.n (n 2+n 2), that is, what is sought;
The second [0,1] 1 sqrt(1+x) dx:
where a=0, b=1, so the integral value.
lim∑ f(k/n)/n
lim∑ (1/sqrt[1+(k/n)])n
lim∑ 1/sqrt(n^2+k*n)
lim 1/sqrt(n^2+1*n)+1/sqrt(n^2+2*n)+.1 sqrt(n 2+n*n), so the proposition is wrong, and each numerator in the sum should be 1 instead of n
According to the expression in the question, when n-> is summed spike or each term in it is ->1, the sum of n terms must be infinite).
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