Advanced Mathematics Definite Integral Problems

If it's 0 to 2, then it's easy to do, using the definite integral department exchange method, first introduce one thing:

2n)!!=(2n)*(2n-2)*(2n-4)*.2,2n-1)!!=(2n-1)*(2n-3)*(2n-5)*.3*1.

The above will be used later, and the following answers are as follows:

sinx) ndx= -(sinx) (n-1)d(cosx), Here we do a commutation first, so the integral variable becomes [1,0], and the above formula is divided into parts:

sinx)^(n-1)d(cosx)=[-(sinx)^(n-1)cosx](1,0)-∫cosxd((sinx)^(n-1))

0-0+∫(n-1)*(cosx)^2*(sinx)^(n-2)dx

n-1)∫[sinx)^(n-2)-(sinx)^n]dx

Shift: n sinx ndx=(n-1) sinx (n-2)dx

Here x is from 0 to 2, iterate repeatedly with the above formula (note the odd and even), and pay attention.

sinxdx=-cosx=1, where x is from 0 to 2, which is equivalent to the original question n=1, sinx 2dx= (1-cos(2x)) 2 dx=[x 2-(sin2x) 4] (0, 2), 4, which is equivalent to the original question n=2, combined with the recursive formula at the beginning to obtain:

n is an odd number, set to n=2k-1, and the original integral value is (2k-2)!!/(2k-1)!!Rule 0!! =1

n is an even number, set to n=2k, and the original integral value is (2k-1)!!/(2k)!!4, where n is not less than 2, and if n=0, then the integral value is obviously 2

If it's from 0 to 2, it will.

There's nothing you can do about getting from A to B.

Summary. Advanced Mathematics Definite Integral Problems

The answer is 1 higher mathematics definite integral problem.

The second question is not 1 2

Analysis] (2axda=c"1}-1, so the answer is: 1【Idea】It can be calculated according to the calculation rule of definite integrals.

Summary. Advanced Mathematics Definite Integral Problems

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Hello, has the question of the foundation been clearly answered for you? If you have any new confusion, please feel free to find me again, and I will answer it for you as soon as possible. I wish you all the best!

1.Observe the upper and lower bounds, the left is x to 1, the right is 1 to 1 x (1 x to 1 in reverse of the ovation), notice that x->1 x, 1->1, can be used as an inverse function.

Settle. So let t=1 x, then x=1 t, dx=-dt t 2 right = [1,1 x].

dx/(1+x^2)=∫1,t]

dt/(t^2

1+1/t^2))=1,t]

dt/(t^2+1)=∫t,1]

dt/(1+t^2)

x,1]dx/(1+x^2)

Note [1,t] denotes a definite integral from 1 to t.

1 is the lower limit and t is the upper limit.

2.Observe the upper and lower bounds, the left side is 0 to , the right chaotic side is 0 to 0 (and vice versa), note that 0-> 0, you can use the function once.

Settle. Let t = -x, then x= -t, dx=-dt left = [0, ].

xf(sinx)dx

-t)f(sint)dt

t)f(sint)dt

f(sint)dt

tf(sint)dt=π∫0,π]

f(sint)dt

xf(sinx)dx

So the book difference is 2 [0, ].

xf(sinx)dx=π∫0,π]

f(sint)dt

So [0, ]

xf(sinx)dx=(π2)∫[0,π]f(sint)dt

f(sinx)dx

13. Let e -x = t, apply the basic integral formula, and you will be.

4. Let x=t, use the partial integration method, you should be able to.

Use the parity of functions to find integrals.

If f(x) is an odd function, then the integral in the symmetric interval (-a,a), (a,a) f(x)dx = 0

If f(x) is an even function, then the integral of (a,a) in the symmetric interval (-a,a), (a,a) f(x)dx =2 (0,a) f(x)dx

Obviously, f(x) = (arcsinx) (1-x) is an even function.

-1/2,1/2) f(x)dx =2 ∫(0,1/2) f(x)dx= 2∫(0,1/2) (arcsinx)²d arcsinx

These questions can be answered directly by applying the basic integral formula and oral arithmetic. You need to strengthen your foundational learning.

newmanhero Mar 27, 2015 19:29:38

An indefinite integral is a known derivative origin function. If f(x)=f(x), then [f(x)+c] = f(x)(c r c is constant).

In other words, integrating f(x) does not necessarily give f(x), because the derivative of f(x)+c is also f(x) (c is an arbitrary constant). So there are an infinite number of results of the f(x) integral, which are indeterminate. We always use f(x)+c instead, which is called an indefinite integral.

That is, if a derivative has primitive functions, then it has an infinite number of primitive functions.

The definite integral is the area of the function f(x) enclosed under the graph line in the interval [a,b]. That is, the area of the graph enclosed by y=0,x=a,x=b,y=f(x). This shape is called a curved trapezoid, with the exception of a curved triangle.

1.Observing the upper and lower bounds, which are x to 1 on the left and 1 to 1 x on the right (1 x to 1 in reverse), note that x->1 x, 1->1, can be solved with an inverse function.

So let t=1 x, then x=1 t, dx=-dt t t 2

Right = [1,1 x] dx (1+x 2) = [1,t] -dt (t 2 * 1+1 t 2)) = [1,t] -dt (t 2+1) = [t,1] dt (1+t 2).

[x,1] dx (1+x 2) Note [1,t] denotes a definite integral from 1 to t, with 1 being the lower bound and t being the upper limit.

2.Observe the upper and lower bounds, 0 to 0 on the left and 0 to 0 on the right (and vice versa), and note that 0-> 0 can be solved with a one-time function.

Let t= -x, then x= -t, dx=-dt

Left = [0, ]xf(sinx)dx = [0] -t)f(sint)dt = [0, ]t) f(sint)dt

∫[0,π]f(sint)dt -∫0,π]tf(sint)dt=π∫[0,π]f(sint)dt -∫0,π]xf(sinx)dx

So 2 [0, ]xf(sinx)dx = [0, ]f(sint)dt

So [0, ]xf(sinx)dx=( 2) [0, ]f(sint)dt =( 2) [0, ]f(sinx)dx

1. Proof: right = (1,1 x) dx (1+x 2) =(let y=1 x, then dx=d(1 y)=-dy y 2) 1,y) -dy y 2*1 [1+(1 y) 2].

(y,1) dy [y 2+1]= (x,1) dx [x 2+1]=left.

2. Proof: Left = (0, )xf(sinx)dx=(let y= -x, then x= -y, dx=-dy) = (0) (y)f[sin( -y)]*dy).

(0,π)y)f(siny)*dy=π∫(0,π)f(siny)*dy-∫(0,π)yf(siny)*dy=π∫(0,π)f(sinx)dx-∫(0,π)xf(sinx)dx

Left = (0, )xf(sinx)dx= 2* (0, )f(sinx)dx=right.

The trick is to substitute.

This one... In the first case, the integral result of 1, (1+x 2) is arctan x.

So, the result of integration from x to 1 is pi 4 - arctan x, and the result of integration from 1 x to 1 is arctan (1 x)-pi 4

Move the item to the right. Arctan x + arctan (1 x) < = PI 2.

Because the result of the definite integral is a number and does not contain letters at all, it has nothing to do with what letters are used; The result of an indefinite integral is a formula that is represented by different letters, such as x2 and t2, which must be different.