Find the definite integral value of the integral e x2 from 0 to 1

The integral e(-x2) is from 0 to 1Definite integralsValue: e (-x 2).Original functionIt's notElementary functionsSo you can't get the points, so this question requires a little skill, and I make f the integral number.

Solve for the square of this value by multiplying two fe (-x 2)dxfe^(-x^2)dx fe^(-x^2)dx=fe^(-x^2)dx fe^(-y^2)dy=ffe^-(x^2+y^2)dxdy，It was this square that became oneDouble integrals, this double integral is easy to calculate, replace the Cartesian coordinate system withPolar coordinate systemCalculate double integrals.

Fundamental theorem. Definite integrals and indefinite integrals seem to be incompatible, but they are intrinsically closely related due to the support of a mathematically important theory. It may seem impossible to infinitely subdivide a graph and add it up, but thanks to this theory, it can be translated into computational integrals.

This theory reveals the connection between the integral and the essence of the Riemann integral, which shows its important position in calculus and even higher mathematics, so the Newton-Leibniz formula is also called the fundamental theorem of calculus.

The original function of e (-x 2) is not an elementary function so you can't integrate, so this problem needs a little skill, I make f an integral sign, first solve the square of this value, that is, take two fe (-x 2) dx multiply, fe (-x 2) dx fe (-x 2) dx = fe (-x 2) dx fe (-y 2)dy=ffe -(x 2+y 2)dxdy, so this square becomes a double integral, this double integral is very easy to calculate, Replacing the Cartesian coordinate system with the polar coordinate system is a double integral, and the fda fre (-r 2)dr = 4 fe (-r 2)dr 2 = 4*-(e -r 2)|(1,0), the result is 4, so this value should be for 4 root number = (1 2) 2

This result is generally expressed as erf(1)*2, and only an approximate value can be obtained, as for erf(x), it is called the error function, which is used in probability theory. For more information on error functions, please see.

Looking downstairs, the integration region [0,1] [0,1] was made wrong after the transformation, and the range of r is actually related to the angle, and it will be separated by 0 4 and 4 2, and the range of r is 0 sec and 0 csc respectively, and the result is still an unusual integral. In fact, (1 2) 2 is the integral value of this integrand from 0 to positive infinity, not the integral value from 0 to 1.

We can do this with respect to the integral of -x squared of e, let f(x)=e (-x 2), g(y)=e (-y 2), and then the equation f(x)*g(y) can be reduced to polar coordinates.

(f(x)*g(y))dxdy= r*e -(r 2)drd, then after the bureau with r, 0 1 points, 0 2 Jitong scatter points, this is the Fang Sell Wild method, you can go to the table to check the results yourself.

e^xdx=e^x+c

So guess the banquet erection, Ding Xiangfeng Jisui Dafen.

0 to 1) e xdx = (e 1 + c) - (e 0 + c) e 1 - e 0e - 1

The answer is E-1

The process of solving the problem is as follows:

-0)lim∑e^(ξi)(△xi)

n-> lim e (i n)(1 n) [where i=i n, xi=1 n,i=1,2,..n】

n->∞lim(1/n)

n->∞lime^(1/n)[1-e]/

n->∞lim[1-e]/

E-1 theoremGeneral theorem.

Theorem 1: Let f(x) be continuous over the interval [a,b], then f(x) is integrable over [a,b].

Theorem 2: If f(x) is bounded by the interval [a,b] and there are only a finite number of discontinuities, then f(x) is integrable on [a,b].

Theorem 3: Let f(x) be monotonic over the interval [a,b], then f(x) is integrable over [a,b].

Common Integration Formulas:

1）∫0dx=c

2）∫x^udx=(x^(u+1))/(u+1)+c

3）∫1/xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

Credits: (0,2)[e x] 2dx

e^x]/2|(0,2)

e 2) big disturbance 2-(e 0) rolling source dan 2

e 2) 2-1 Crack do 2

Answer: 1 [1+e (x-1)] dx

1+e^(x-1)-e^(x-1)]/1+e^(x-1)] dx∫1-e^(x-1)/[1+e^(x-1)] dxx-∫1/[1+e^(x-1)] d[1+e^(x-1)]x-ln[1+e^(x-1)] c

So the definite integral (0 to 1)1 [1+e (x-1)] dxx-ln[1+e (x-1)] 0 to 1).

1-ln(1+e^0)-0+ln(1+e^(-1))1-ln(2+2/e)