- #1
cd246
- 30
- 0
Homework Statement
/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2
Homework Equations
The Attempt at a Solution
u^-1/2*(1/2)du
(-1/2)(u^1/2)
(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1
cd246 said:Homework Statement
/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2
Homework Equations
The Attempt at a Solution
u^-1/2*(1/2)du
(-1/2)(u^1/2)
(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1
Well, no, that is not the answer! The answer is (2t+4)1/2+ C where C can be any constant. Is it possible that your integral had limits that you haven't told us?cd246 said:Homework Statement
/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2
Please, please, pleas, tell us what you are doing! This makes no sense until you tell us that you are using the substitution u= 2x- 4 so that (2x-4)-1/2 becomes u-1/2 (not 1/2) and 2dx= du so dx= (1/2)duHomework Equations
The Attempt at a Solution
u^-1/2*(1/2)du
?? The anti-derivative of un is 1/(n+1) un+1 (as long as n is not -1). For n= -1/2, n+1= 1/2 and that becomes 2 u1/2, Multiplying by 1/2 from the "dx= (1/2) du" gives u1/2+ C= (2t+4)1/2+ C.(-1/2)(u^1/2)
(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1
HallsofIvy said:Well, no, that is not the answer! The answer is (2t+4)1/2+ C where C can be any constant. Is it possible that your integral had limits that you haven't told us?
Please, please, pleas, tell us what you are doing! This makes no sense until you tell us that you are using the substitution u= 2x- 4 so that (2x-4)-1/2 becomes u-1/2 (not 1/2) and 2dx= du so dx= (1/2)du
?? The anti-derivative of un is 1/(n+1) un+1 (as long as n is not -1). For n= -1/2, n+1= 1/2 and that becomes 2 u1/2, Multiplying by 1/2 from the "dx= (1/2) du" gives u1/2+ C= (2t+4)1/2+ C.
is that for simple linear substitutions, ax+ b, you should be able to just think: do the integral ignoring the linear part inside the power, and the divide by a: The anti-derivative of (ax+ b)n is (n/a)(ax+ b)n+1[/sup]. You don't really need to write out the substitution. That comes with experience.rootX said:-You don't need substitution for this.
simply use this exp:
[tex]\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}[/tex]
>also recheck your calculations.
Substitution with integration is a technique used in calculus to simplify the process of integration. It involves substituting a variable or expression in the integral with a new variable or expression in order to make the integral easier to solve.
The substitution to use is determined by looking for a part of the integral that resembles a known integral. This part is then substituted with a new variable, chosen in a way that simplifies the integral.
The general formula for substitution with integration is: ∫f(u) du = ∫f(g(x))g'(x) dx, where u = g(x) and du = g'(x) dx.
No, substitution with integration can only be used to solve integrals that can be transformed into a known integral by substituting a variable or expression.
Yes, substitution with integration can be used for definite integrals. The limits of integration must also be transformed when substituting the variable or expression in order to obtain the correct result.