definite integral

[Primarygun]

Here I got a conceptual problem rarely mentioned by my teacher.

When we try to figure the value of a definite integral, the general method is to find out the expression(value) of the indefinite integral, right?

but, when we change the dx to d(-x), a must change must be produced correspondingly, that's changing the upper limit and lower limit, right?

 

[math]

\int_a^b x \,dx

[/math]

[matt grime]

since these are dummy variables let's keep the names separate.

 

so set y=-x, and dy=-dx

 

then the upper limit of x=b must mean y=-x=-b, and the lower limit becomes -a.

[Primarygun]

Thanks for kind reply to my question.

[math]

\int_a^b f(x) \,dx =

\int_a^b f(y) \,dy

[/math]

This is where a dummy variable could be used.

But, put y=-x and f(x)=x^2, from the formula above, We must keep the limits constant, However, this gives a different answer.Why?

[matt grime]

The substitution y=-x gives the same answer. Just try it. It changes the integral to

 

[math]\int_{-a}^{-b}-y^2 dy[/math]

 

Why do you think you 'must keep the limits constant' (which doesn't really make sense).

[Primarygun]

couldn't y be the dummy variable?

Thanks

[matt grime]

y is a dummy variable, so is x, that is why you can do substitutions.

[Primarygun]

If I put -x as the dummy variable, it is correct?

i.e. replace the y with -x directly without changing the limit

[Connor]

no, because you must substitute the dummy for the limits too

[Primarygun]

no, because you must substitute the dummy for the limits too

Why?

Isn't the formula in reply 3 true?

[matt grime]

What are the limits they go from x=a to x=b, so if you set y=-x, then why are the limits y=a to y=b the same after this subsitution? You've just asserted that a=x=-y=-a and b=-b similarly.

 

Certainly the formula in post 3 is true, but it is substituting y=x.

 

Doing y=-x does not effect that change of integral, when f(x)=x^2, the substitution also changes the dx to a -dy, doesn't it?

 

Moral of story: don't mix and match, changing x to -y must be done at all points the x appears.

 

The fact that

 

[math]\int_a^bx^2dx=\int_a^b(-x)^2dx[/math]

 

does not happen because of a substitution, it is because x^2=(-x)^2

[Primarygun]

Doing y=-x does not effect that change of integral, when f(x)=x^2, the substitution also changes the dx to a -dy, doesn't it?

When can we use the formula in reply 3?

[matt grime]

You don't give a formula in post three apart from one that is trivially always true

 

[math]\int_a^bf(x)dx=\int_a^bf(y)dy[/math]

 

that is gotten by subsituting y=x in to the integral.

[Primarygun]

Apart from relating y=x, what else can x be substituted?

[Dave]

You can set x to be equal to (pretty much) anything as long as you change the limits. A typical substitution might be something like [imath]x = \cos\theta[/imath].

[Connor]

okay, I guess you don't have to change the limits if you substitute the original variable back after integration

[matt grime]

Apart from relating y=x, what else can x be substituted?

 

Absolutely any other letter, or any other reasonable function of x (ie don't pick one that is not differentiable).

 

Now, what was the point of all this?