Integration by substitution











In calculus, integration by substitution, also known as u-substitution, is a method for finding integrals. Using the fundamental theorem of calculus often requires finding an antiderivative. For this and other reasons, integration by substitution is an important tool in mathematics. It is the counterpart to the chain rule for differentiation.




Contents





  • 1 Substitution for a single variable

    • 1.1 Proposition


    • 1.2 Proof


    • 1.3 Examples

      • 1.3.1 Example 1: from right to left


      • 1.3.2 Example 2: from left to right


      • 1.3.3 Example 3: antiderivatives




  • 2 Substitution for multiple variables


  • 3 Application in probability


  • 4 See also


  • 5 References


  • 6 External links




Substitution for a single variable



Proposition


Let IR be an interval and φ : [a,b] → I be a differentiable function with integrable derivative. Suppose that f : IR is a continuous function. Then


∫φ(a)φ(b)f(u)du=∫abf(φ(x))φ′(x)dx.displaystyle int _varphi (a)^varphi (b)f(u),du=int _a^bf(varphi (x))varphi '(x),dx.displaystyle int _varphi (a)^varphi (b)f(u),du=int _a^bf(varphi (x))varphi '(x),dx.

In Leibniz notation, the substitution u = φ(x) yields


dudx=φ′(x).displaystyle frac dudx=varphi '(x).displaystyle frac dudx=varphi '(x).

Working heuristically with infinitesimals yields the equation


du=φ′(x)dx,displaystyle du=varphi '(x),dx,displaystyle du=varphi '(x),dx,

which suggests the substitution formula above. (This equation may be put on a rigorous foundation by interpreting it as a statement about differential forms.) One may view the method of integration by substitution as a partial justification of Leibniz's notation for integrals and derivatives.


The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be used from left to right or from right to left in order to simplify a given integral. When used in the latter manner, it is sometimes known as u-substitution or w-substitution.



Proof


Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let f and φ be two functions satisfying the above hypothesis that f is continuous on I and φ is integrable on the closed interval [a,b]. Then the function f(φ(x))φ′(x) is also integrable on [a,b]. Hence the integrals


∫φ(a)φ(b)f(u)dudisplaystyle int _varphi (a)^varphi (b)f(u),dudisplaystyle int _varphi (a)^varphi (b)f(u),du

and


∫abf(φ(x))φ′(x)dxdisplaystyle int _a^bf(varphi (x))varphi '(x),dxdisplaystyle int _a^bf(varphi (x))varphi '(x),dx

in fact exist, and it remains to show that they are equal.


Since f is continuous, it has an antiderivative F. The composite function Fφ is then defined. Since φ is differentiable, combining the chain rule and the definition of an antiderivative gives


(F∘φ)′(x)=F′(φ(x))φ′(x)=f(φ(x))φ′(x).displaystyle (Fcirc varphi )'(x)=F'(varphi (x))varphi '(x)=f(varphi (x))varphi '(x).displaystyle (Fcirc varphi )'(x)=F'(varphi (x))varphi '(x)=f(varphi (x))varphi '(x).

Applying the fundamental theorem of calculus twice gives


∫abf(φ(x))φ′(x)dx=∫ab(F∘φ)′(x)dx=(F∘φ)(b)−(F∘φ)(a)=F(φ(b))−F(φ(a))=∫φ(a)φ(b)f(u)du,displaystyle beginalignedint _a^bf(varphi (x))varphi '(x),dx&=int _a^b(Fcirc varphi )'(x),dx\&=(Fcirc varphi )(b)-(Fcirc varphi )(a)\&=F(varphi (b))-F(varphi (a))\&=int _varphi (a)^varphi (b)f(u),du,endaligneddisplaystyle beginalignedint _a^bf(varphi (x))varphi '(x),dx&=int _a^b(Fcirc varphi )'(x),dx\&=(Fcirc varphi )(b)-(Fcirc varphi )(a)\&=F(varphi (b))-F(varphi (a))\&=int _varphi (a)^varphi (b)f(u),du,endaligned

which is the substitution rule.



Examples



Example 1: from right to left


Consider the integral


∫02xcos⁡(x2+1)dx.displaystyle int _0^2xcos(x^2+1),dx.displaystyle int _0^2xcos(x^2+1),dx.

If we apply the formula from right to left and make the substitution u = φ(x) = x2 + 1, we obtain du = 2x dx and hence x dx = ½du. Therefore


∫x=0x=2xcos⁡(x2+1)dx=12∫u=1u=5cos⁡(u)du=12(sin⁡(5)−sin⁡(1)).displaystyle beginalignedint _x=0^x=2xcos(x^2+1),dx&=frac 12int _u=1^u=5cos(u),du\&=frac 12(sin(5)-sin(1)).endaligneddisplaystyle beginalignedint _x=0^x=2xcos(x^2+1),dx&=frac 12int _u=1^u=5cos(u),du\&=frac 12(sin(5)-sin(1)).endaligned

Since the lower limit x = 0 was replaced with u = 02 + 1 = 1, and the upper limit x = 2 replaced with u = 22 + 1 = 5, a transformation back into terms of x was unnecessary.



Example 2: from left to right


For the integral


∫011−x2dx,displaystyle int _0^1sqrt 1-x^2,dx,displaystyle int _0^1sqrt 1-x^2,dx,

the formula needs to be used from left to right. The substitution x = sin(u), dx = cos(u) du is useful because 1−sin2⁡u=cos⁡(u)displaystyle sqrt 1-sin ^2u=cos(u)displaystyle sqrt 1-sin ^2u=cos(u):


∫011−x2dx=∫0π/21−sin2⁡ucos⁡(u)du=∫0π/2cos2⁡udu=(u2+sin⁡(2u)4)|0π/2=π4+0=π4.displaystyle beginalignedint _0^1sqrt 1-x^2,dx&=int _0^pi /2sqrt 1-sin ^2ucos(u),du\&=int _0^pi /2cos ^2u,du\&=left(frac u2+frac sin(2u)4right)Bigg vert _0^pi /2\&=frac pi 4+0=frac pi 4.endaligneddisplaystyle beginalignedint _0^1sqrt 1-x^2,dx&=int _0^pi /2sqrt 1-sin ^2ucos(u),du\&=int _0^pi /2cos ^2u,du\&=left(frac u2+frac sin(2u)4right)Bigg vert _0^pi /2\&=frac pi 4+0=frac pi 4.endaligned

The resulting integral can be computed using integration by parts or a double angle formula followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or π / 4.



Example 3: antiderivatives


Substitution can be used to determine antiderivatives. One chooses a relation between x and u, determines the corresponding relation between dx and du by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between u and x is then undone.


Similar to our first example above, we can determine the following antiderivative with this method:


∫xcos⁡(x2+1)dx=12∫2xcos⁡(x2+1)dx=12∫cos⁡udu=12sin⁡u+C=12sin⁡(x2+1)+C,displaystyle beginalignedint xcos(x^2+1),dx&=frac 12int 2xcos(x^2+1),dx\&=frac 12int cos u,du\&=frac 12sin u+C=frac 12sin(x^2+1)+C,endaligneddisplaystyle beginalignedint xcos(x^2+1),dx&=frac 12int 2xcos(x^2+1),dx\&=frac 12int cos u,du\&=frac 12sin u+C=frac 12sin(x^2+1)+C,endaligned

where C is an arbitrary constant of integration.


Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution u = x2 + 1.



Substitution for multiple variables


One may also use substitution when integrating functions of several variables.
Here the substitution function (v1,...,vn) = φ(u1, ..., un) needs to be injective and continuously differentiable, and the differentials transform as


dv1⋯dvn=|det(Dϕ)(u1,…,un)|du1⋯dun,displaystyle dv_1cdots dv_n=displaystyle dv_1cdots dv_n=

where det()(u1, ..., un) denotes the determinant of the Jacobian matrix of partial derivatives of φ at the point (u1, ..., un). This formula expresses the fact that the absolute value of the determinant of a matrix equals the volume of the parallelotope spanned by its columns or rows.


More precisely, the change of variables formula is stated in the next theorem:


Theorem. Let U be an open set in Rn and φ : URn an injective differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every x in U. Then for any real-valued, compactly supported, continuous function f, with support contained in φ(U),


∫φ(U)f(v)dv=∫Uf(φ(u))|det(Dφ)(u)|du.det(Dvarphi )(mathbf u )rightdet(Dvarphi )(mathbf u )right

The conditions on the theorem can be weakened in various ways. First, the requirement that φ be continuously differentiable can be replaced by the weaker assumption that φ be merely differentiable and have a continuous inverse (Rudin 1987, Theorem 7.26). This is guaranteed to hold if φ is continuously differentiable by the inverse function theorem. Alternatively, the requirement that det() ≠ 0 can be eliminated by applying Sard's theorem (Spivak 1965).


For Lebesgue measurable functions, the theorem can be stated in the following form (Fremlin 2010, Theorem 263D):


Theorem. Let U be a measurable subset of Rn and φ : URn an injective function, and suppose for every x in U there exists φ′(x) in Rn,n such that φ(y) = φ(x) + φ′(x)(yx) + o(||yx||) as yx (here o is little-o notation). Then φ(U) is measurable, and for any real-valued function f defined on φ(U),


∫φ(U)f(v)dv=∫Uf(φ(u))|detφ′(u)|dudisplaystyle int _varphi (U)f(v),dv=int _Uf(varphi (u))leftdisplaystyle int _varphi (U)f(v),dv=int _Uf(varphi (u))left

in the sense that if either integral exists (including the possibility of being properly infinite), then so does the other one, and they have the same value.


Another very general version in measure theory is the following (Hewitt & Stromberg 1965, Theorem 20.3):


Theorem. Let X be a locally compact Hausdorff space equipped with a finite Radon measure μ, and let Y be a σ-compact Hausdorff space with a σ-finite Radon measure ρ. Let φ : XY be a continuous and absolutely continuous function (where the latter means that ρ(φ(E)) = 0 whenever μ(E) = 0). Then there exists a real-valued Borel measurable function w on X such that for every Lebesgue integrable function f : YR, the function (fφ) ⋅ w is Lebesgue integrable on X, and


∫Yf(y)dρ(y)=∫X(f∘φ)(x)w(x)dμ(x).displaystyle int _Yf(y),drho (y)=int _X(fcirc varphi )(x),w(x),dmu (x).displaystyle int _Yf(y),drho (y)=int _X(fcirc varphi )(x),w(x),dmu (x).

Furthermore, it is possible to write


w(x)=(g∘φ)(x)displaystyle w(x)=(gcirc varphi )(x)displaystyle w(x)=(gcirc varphi )(x)

for some Borel measurable function g on Y.


In geometric measure theory, integration by substitution is used with Lipschitz functions. A bi-Lipschitz function is a Lipschitz function φ : URn which is injective and whose inverse function φ−1 : φ(U) → U is also Lipschitz. By Rademacher's theorem a bi-Lipschitz mapping is differentiable almost everywhere. In particular, the Jacobian determinant of a bi-Lipschitz mapping det is well-defined almost everywhere. The following result then holds:


Theorem. Let U be an open subset of Rn and φ : URn be a bi-Lipschitz mapping. Let f : φ(U) → R be measurable. Then


∫U(f∘φ)(x)|detDφ(x)|dx=∫φ(U)f(x)dx,dx=int _varphi (U)f(x),dx,dx=int _varphi (U)f(x),dx

in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.


The above theorem was first proposed by Euler when he developed the notion of double integrals in 1769. Although generalized to triple integrals by Lagrange in 1773, and used by Legendre, Laplace, Gauss, and first generalized to n variables by Mikhail Ostrogradski in 1836, it resisted a fully rigorous formal proof for a surprisingly long time, and was first satisfactorily resolved 125 years later, by Élie Cartan in a series of papers beginning in the mid-1890s (Katz 1982; Ferzola 1994).



Application in probability


Substitution can be used to answer the following important question in probability: given a random variable Xdisplaystyle XX with probability density pXdisplaystyle p_Xp_X and another random variable Ydisplaystyle YY related to Xdisplaystyle XX by the equation y=ϕ(x)displaystyle y=phi (x)y=phi (x), what is the probability density for Ydisplaystyle YY?


It is easiest to answer this question by first answering a slightly different question: what is the probability that Ydisplaystyle YY takes a value in some particular subset Sdisplaystyle SS? Denote this probability P(Y∈S)displaystyle P(Yin S)P(Yin S). Of course, if Ydisplaystyle YY has probability density pYdisplaystyle p_Yp_Y then the answer is


P(Y∈S)=∫SpY(y)dy,displaystyle P(Yin S)=int _Sp_Y(y),dy,displaystyle P(Yin S)=int _Sp_Y(y),dy,

but this isn't really useful because we don't know pYdisplaystyle p_Yp_Y; it's what we're trying to find. We can make progress by considering the problem in the variable Xdisplaystyle XX. Ydisplaystyle YY takes a value in Sdisplaystyle SS whenever Xdisplaystyle XX takes a value in ϕ−1(S)displaystyle phi ^-1(S)phi ^-1(S), so


P(Y∈S)=∫ϕ−1(S)pX(x)dx.displaystyle P(Yin S)=int _phi ^-1(S)p_X(x),dx.displaystyle P(Yin S)=int _phi ^-1(S)p_X(x),dx.

Changing from variable xdisplaystyle xx to ydisplaystyle yy gives


P(Y∈S)=∫ϕ−1(S)pX(x)dx=∫SpX(ϕ−1(y))|dϕ−1dy|dy.displaystyle P(Yin S)=int _phi ^-1(S)p_X(x),dx=int _Sp_X(phi ^-1(y))leftdisplaystyle P(Yin S)=int _phi ^-1(S)p_X(x),dx=int _Sp_X(phi ^-1(y))left

Combining this with our first equation gives


∫SpY(y)dy=∫SpX(ϕ−1(y))|dϕ−1dy|dy,frac dphi ^-1dyrightfrac dphi ^-1dyright

so


pY(y)=pX(ϕ−1(y))|dϕ−1dy|.displaystyle p_Y(y)=p_X(phi ^-1(y))leftdisplaystyle p_Y(y)=p_X(phi ^-1(y))left

In the case where Xdisplaystyle XX and Ydisplaystyle YY depend on several uncorrelated variables, i.e. pX=pX(x1,…,xn)displaystyle p_X=p_X(x_1,ldots ,x_n)displaystyle p_X=p_X(x_1,ldots ,x_n) and y=ϕ(x)displaystyle y=phi (x)y=phi (x), pYdisplaystyle p_Yp_Y can be found by substitution in several variables discussed above. The result is


pY(y)=pX(ϕ−1(y))|detDϕ−1(y)|.det Dphi ^-1(y)rightdet Dphi ^-1(y)right


See also


  • Probability density function

  • Substitution of variables

  • Tangent half-angle substitution

  • Trigonometric substitution


References



  • Ferzola, Anthony P. (1994), "Euler and differentials", The College Mathematics Journal, 25 (2): 102&ndash, 111, doi:10.2307/2687130.mw-parser-output cite.citationfont-style:inherit.mw-parser-output qquotes:"""""""'""'".mw-parser-output code.cs1-codecolor:inherit;background:inherit;border:inherit;padding:inherit.mw-parser-output .cs1-lock-free abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-subscription abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registrationcolor:#555.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration spanborder-bottom:1px dotted;cursor:help.mw-parser-output .cs1-hidden-errordisplay:none;font-size:100%.mw-parser-output .cs1-visible-errorfont-size:100%.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-formatfont-size:95%.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-leftpadding-left:0.2em.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-rightpadding-right:0.2em


  • Fremlin, D.H. (2010), Measure Theory, Volume 2, Torres Fremlin, ISBN 978-0-9538129-7-4.


  • Hewitt, Edwin; Stromberg, Karl (1965), Real and Abstract Analysis, Springer-Verlag, ISBN 978-0-387-04559-7.


  • Katz, V. (1982), "Change of variables in multiple integrals: Euler to Cartan", Mathematics Magazine, 55 (1): 3&ndash, 11, doi:10.2307/2689856


  • Rudin, Walter (1987), Real and Complex Analysis, McGraw-Hill, ISBN 978-0-07-054234-1.


  • Spivak, Michael (1965), Calculus on Manifolds, Westview Press, ISBN 978-0-8053-9021-6.


External links







  • Integration by substitution at Encyclopedia of Mathematics


  • Area formula at Encyclopedia of Mathematics


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