Integration by substitution
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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 I ⊆ R be an interval and φ : [a,b] → I be a differentiable function with integrable derivative. Suppose that f : I → R 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.
In Leibniz notation, the substitution u = φ(x) yields
- dudx=φ′(x).displaystyle frac dudx=varphi '(x).
Working heuristically with infinitesimals yields the equation
- du=φ′(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),du
and
- ∫abf(φ(x))φ′(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 F ∘ φ 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).
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,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.
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)).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,
the formula needs to be used from left to right. The substitution x = sin(u), dx = cos(u) du is useful because 1−sin2u=cos(u)displaystyle sqrt 1-sin ^2u=cos(u):
- ∫011−x2dx=∫0π/21−sin2ucos(u)du=∫0π/2cos2udu=(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.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∫cosudu=12sinu+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,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=
where det(Dφ)(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 : U → Rn 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 )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(Dφ) ≠ 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 : U → Rn 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)(y − x) + o(||y − x||) as y → x (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))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 : X → Y 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 : Y → R, the function (f ∘ φ) ⋅ 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).
Furthermore, it is possible to write
- w(x)=(g∘φ)(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 : U → Rn 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 Dφ is well-defined almost everywhere. The following result then holds:
Theorem. Let U be an open subset of Rn and φ : U → Rn 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
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 X with probability density pXdisplaystyle p_X and another random variable Ydisplaystyle Y related to Xdisplaystyle X by the equation y=ϕ(x)displaystyle y=phi (x), what is the probability density for Ydisplaystyle Y?
It is easiest to answer this question by first answering a slightly different question: what is the probability that Ydisplaystyle Y takes a value in some particular subset Sdisplaystyle S? Denote this probability P(Y∈S)displaystyle P(Yin S). Of course, if Ydisplaystyle Y has probability density pYdisplaystyle p_Y then the answer is
- P(Y∈S)=∫SpY(y)dy,displaystyle P(Yin S)=int _Sp_Y(y),dy,
but this isn't really useful because we don't know pYdisplaystyle p_Y; it's what we're trying to find. We can make progress by considering the problem in the variable Xdisplaystyle X. Ydisplaystyle Y takes a value in Sdisplaystyle S whenever Xdisplaystyle X takes a value in ϕ−1(S)displaystyle 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.
Changing from variable xdisplaystyle x to ydisplaystyle y 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))left
Combining this with our first equation gives
- ∫SpY(y)dy=∫SpX(ϕ−1(y))|dϕ−1dy|dy,frac dphi ^-1dyright
so
- pY(y)=pX(ϕ−1(y))|dϕ−1dy|.displaystyle p_Y(y)=p_X(phi ^-1(y))left
In the case where Xdisplaystyle X and Ydisplaystyle Y depend on several uncorrelated variables, i.e. pX=pX(x1,…,xn)displaystyle p_X=p_X(x_1,ldots ,x_n) and y=ϕ(x)displaystyle y=phi (x), pYdisplaystyle p_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)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
The Wikibook Calculus has a page on the topic of: The Substitution Rule |
Wikiversity has learning resources about Integration by Substitution |
Integration by substitution at Encyclopedia of Mathematics
Area formula at Encyclopedia of Mathematics