Substitution

Substitution for Integrals

Substitution Rule for Indefinite Integrals

If \(u=g(x)\) is a differential function whose range is an interval \(I\), and \(f\) is continuous on \(I\), then

\[\int f(g(x))\cdot {g}'(x)dx = \int f(u)du\]

Proof

By the Chain Rule, \(F(g(x))\) is an antiderivative of \(f(g(x))\cdot {g}'(x)\) whenever \(F\) is an antiderivative of \(f\), beacuse

\[\begin{split}\begin{align} \cfrac{d}{dx}F(g(x)) & = {F}'(g(x))\cdot {g}'(x)\\ & = f(g(x)){g}'(x) \end{align}\end{split}\]

If we make the substitution \(u=g(x)\), then

\[\begin{split}\begin{align} \int f(g(x))\cdot {g}'(x)dx &= \int \cfrac{d}{dx}F(g(x))dx\\ &=F(g(x))+C\\ &=F(u)+C\\ &=\int {F}'(u)du\\ &=\int f(u)du\\ \end{align}\end{split}\]

Substitution In Definite Integrals

If \({g}'\) is continuous on the interval \([a,b]\) and \(f\) is continuous on the range of \(g(x)=u\), then

\[\begin{split}\int_{a}^{b}f(g(x))\cdot {g}'(x)dx=\int_{g(a)}^{g(b)}f(u)du\\\end{split}\]

Proof

Let \(F\) denote any antiderivative of \(f\). Then,

\[\begin{split}\begin{aligned} \int_{a}^{b} f(g(x)) \cdot g^{\prime}(x) d x & =F(g(x))\Bigg ]_{x=a}^{x=b} \\ & =F(g(b))-F(g(a)) \\ & =F(u)\Bigg]_{u=g(a)}^{u=g(b)} \\ & =\int_{g(a)}^{g(b)} f(u) d u . \end{aligned}\end{split}\]

Substitution Rule for Indefinite Integrals( version 2)

For convenience, swap \(x\) and \(u\)

If \(x=g(u)\) is a differential function whose range is an interval \(I\), and \(f\) is continuous on \(I\), then

\[\int f(x)dx = \int f(g(u))\cdot {g}'(u)du\]

Proof

By the Chain Rule, \(F(g(u))\) is an antiderivative of \(f(g(u))\cdot {g}'(u)\) whenever \(F\) is an antiderivative of \(f\), beacuse

\[\begin{split}\begin{align} \cfrac{d}{du}F(g(u)) & = {F}'(g(u))\cdot {g}'(u)\\ & = f(g(u)){g}'(u) \end{align}\end{split}\]

If we make the substitution \(x=g(u)\), then

\[\begin{split}\begin{align} \int f(g(u))\cdot {g}'(u)du &= \int \cfrac{d}{du}F(g(u))du\\ &=F(g(u))+C\\ &=F(x)+C\\ &=\int {F}'(x)dx\\ &=\int f(x)dx\\ \end{align}\end{split}\]

Substitution In Definite Integrals( version 2)

If \({g}'\) is continuous on the interval \([a,b]\) and \(f\) is continuous on the range of \(g(u)=x\), then

\[\int_{g(a)}^{g(b)}f(x)dx=\int_{a}^{b}f(g(u))\cdot {g}'(u)du\]

Proof

Let \(F\) denote any antiderivative of \(f\). Then,

\[\begin{split}\begin{aligned} \int_{a}^{b} f(g(x)) \cdot g^{\prime}(x) d x & =F(g(x))\Bigg ]_{x=a}^{x=b} \\ & =F(g(b))-F(g(a)) \\ & =F(u)\Bigg]_{u=g(a)}^{u=g(b)} \\ & =\int_{g(a)}^{g(b)} f(u) d u . \end{aligned}\end{split}\]

Some Example

\[\int_{g(a)}^{g(b)}f(x)dx=\int_{a}^{b}f(g(u)){g}'(u) du\]

\[\begin{split}x=g(u)\\\end{split}\]

\[\begin{split}\left\{\begin{matrix} u=a,& x=g(a)\\ u=b,& x=g(b)\\ \end{matrix}\right.\end{split}\]

\[\begin{split}\int_{g(a)}^{g(b)}f(x)dx=\int_{a}^{b}f(g(u))J(u) du\\\end{split}\]

\[\begin{split}J(u)=\text{det}\begin{bmatrix} \cfrac{\partial x}{\partial u} \end{bmatrix}=\text{det}\begin{bmatrix} \cfrac{\partial g}{\partial u} \end{bmatrix}=\text{det}\begin{bmatrix} \cfrac{\text{d} g}{\text{d} u} \end{bmatrix}=\text{det}\begin{bmatrix} {g}'(u) \end{bmatrix}={g}'(u)\\\end{split}\]

Example 1:

\[\begin{split}x=g(u,t)=(1-u)t+\frac{1}{2}ut^2+u \\\end{split}\]

\[\begin{split}\left\{\begin{matrix} u=0,t=1,& x=g(u,t)=g(0,1)=1\\ u=1,t=1,& x=g(u,t)=g(1,1)=\frac{1}{2}+1\\ \end{matrix}\right.\end{split}\]

Let

\[f(x,t)\equiv 1\]

then there is

\[\int_{g(a,t)}^{g(b,t)}f(x,t)dx=\int_{1}^{1+\frac{1}{2} }1dx=\frac{1}{2}\]

\[\cfrac{\partial g(u,t)}{\partial u}=\cfrac{\partial [(1-u)t+\frac{1}{2}ut^2+u]}{\partial u}=-t+\frac{1}{2}t^2+1\]

\[\cfrac{\partial g(u,t)}{\partial u}\Bigg|_{t=1}=\cfrac{\partial [(1-u)t+\frac{1}{2}ut^2+u]}{\partial u}\Bigg|_{t=1}=(-t+\frac{1}{2}t^2+1)\Bigg|_{t=1}=\frac{1}{2}\]

\[\begin{split}\int_{g(a,t)}^{g(b,t)}f(x,t)dx=\int_{a(t)}^{b(t)}f(g(u,t),t)\cfrac{\partial g(u,t)}{\partial u} du\\\end{split}\]

\[\int_{a(t)}^{b(t)}f(g(u,t),t)\cfrac{\partial g(u,t)}{\partial u} du=\int_{0}^{1}1\times\frac{1}{2} du=\frac{1}{2}\]

Let

\[f(x,t)\equiv x\]

then there is

\[\int_{g(a,t)}^{g(b,t)}f(x,t)dx=\int_{1}^{1+\frac{1}{2} }xdx=\frac{1}{2}x^2\Bigg|_{1}^{1+\frac{1}{2}} =\frac{1}{2}((\frac{3}{2})^2-1^{2})=\frac{5}{8}\]

\[\cfrac{\partial g(u,t)}{\partial u}\Bigg|_{t=1}=\cfrac{\partial [(1-u)t+\frac{1}{2}ut^2+u]}{\partial u}\Bigg|_{t=1}=(-t+\frac{1}{2}t^2+1)\Bigg|_{t=1}=\frac{1}{2}\]

\[\int_{g(a,t)}^{g(b,t)}f(x,t)dx=\int_{a(t)}^{b(t)}f(g(u,t),t)\cfrac{\partial g(u,t)}{\partial u} du\]

\[\begin{split}x=g(u,t)=(1-u)t+\frac{1}{2}ut^2+u \\\end{split}\]

\[\begin{split}\begin{align} \displaystyle \int_{a(t)}^{b(t)}f(g(u,t),t)\cfrac{\partial g(u,t)}{\partial u} du & = \int_{0}^{1}((1-u)1+\frac{1}{2}u1^2+u)\times\frac{1}{2} du\\ \displaystyle & = \int_{0}^{1}(1+\frac{1}{2}u)\times\frac{1}{2} du\\ \displaystyle& = \frac{1}{2}(u+\frac{1}{4}u^2)\Bigg|_{0}^{1}\\ \displaystyle& = \frac{1}{2}(1+\frac{1}{4})\\ & = \frac{5}{8} \end{align}\end{split}\]

Example 2:

\[\begin{split}\begin{array}{l} x=g(u,t)\\ f(x,t)=f(g(u,t),t)=\hat{f}(u,t)\\ \cfrac{\text{d}\hat{f}(u,t)}{\text{d}t}\equiv \cfrac{\partial \hat{f}(u,t)}{\partial t} =\cfrac{\partial {f}(x,t)}{\partial t}+\cfrac{\partial {f}(x,t)}{\partial x}\cfrac{\partial {g}(u,t)}{\partial t}\\ \cfrac{\text{d}\hat{f}}{\text{d}t}\equiv \cfrac{\partial \hat{f}}{\partial t} =\cfrac{\partial {f}}{\partial t}+\cfrac{\partial {f}}{\partial x}\cfrac{\partial {g}}{\partial t} \end{array}\end{split}\]

\[\begin{split}x=g(u,t)=(1-u)t+\frac{1}{2}ut^2+u \\\end{split}\]

\[\begin{split}\begin{array}{l} f(x,t)= xt \\ f(x,t)= xt=((1-u)t+\frac{1}{2}ut^2+u)t=\hat{f}(u,t) \\ \hat{f}(u,t)=((1-u)t+\frac{1}{2}ut^2+u)t=((1-u)t^2+\frac{1}{2}ut^3+ut) \end{array}\end{split}\]

\[\cfrac{\text{d}\hat{f}(u,t)}{\partial t}\equiv\cfrac{\partial \hat{f}(u,t)}{\partial t} =\cfrac{\partial ((1-u)t^2+\cfrac{1}{2}ut^3+ut)}{\partial t} =(2(1-u)t+3\cfrac{1}{2}ut^2+u)\]

\[\begin{split}\cfrac{\partial f(x,t)}{\partial t}=x,\quad\cfrac{\partial f(x,t)}{\partial x}=t\\\end{split}\]

\[\begin{split}\begin{align} \cfrac{\partial x}{\partial t} & = \cfrac{\partial g(u,t)}{\partial t}\\ & = \cfrac{\partial ((1-u)t+\frac{1}{2}ut^2+u)}{\partial t}\\ & = (1-u)+ut \end{align}\end{split}\]

\[\begin{split}\begin{align} \cfrac{\text{d}f(x,t)}{\text{d} t} & = \cfrac{\partial {f}(x,t)}{\partial t}+\cfrac{\partial {f}(x,t)}{\partial x}\cfrac{\partial {g}(u,t)}{\partial t}\\ & = x+t((1-u)+ut)\\ & = x+((1-u)t+ut^2)\\ & = ((1-u)t+\frac{1}{2}ut^2+u)+((1-u)t+ut^2)\\ & = (2(1-u)t+\frac{3}{2}ut^2+u)\\ \end{align}\end{split}\]

A better understanding of this formula

\[\begin{split}\begin{array}{c} x=g(u,t)=(1-u)t+\cfrac{1}{2}ut^2+u\\ {g}'(u,t)=\cfrac{\partial g(u,t)}{\partial u}=-t+\cfrac{1}{2}t^2+1\\ \displaystyle \int_{g(a)}^{g(b)}\cfrac{d f(x,t)}{dt} dx=\int_{a}^{b}\cfrac{\partial \hat{f}(u,t)}{\partial t}{g}'(u,t) du \end{array}\end{split}\]

Proof:

\[\begin{split}\begin{array}{c} \displaystyle h(x,t)\equiv \cfrac{d f(x,t)}{dt}\\ \displaystyle \hat{h}(u,t)\equiv \cfrac{\partial \hat{f}(u,t)}{\partial t}\\ \end{array}\end{split}\]

\[h(x,t) = \hat{h}(u,t)\]

By the Chain Rule, \(F(g(u))\) is an antiderivative of \(h(g(u))\cdot {g}'(u)\) whenever \(F\) is an antiderivative of \(h\), beacuse

\[\begin{split}\begin{align} \cfrac{d}{du}F(g(u)) & = {F}'(g(u))\cdot {g}'(u)\\ & = h(g(u)){g}'(u) \end{align}\end{split}\]

If we make the substitution \(x=g(u,t)\), then

\[\begin{split}\begin{align} \int h(g(u))\cdot {g}'(u)du &= \int \cfrac{d}{du}F(g(u))du\\ &=F(g(u))+C\\ &=F(x)+C\\ &=\int {F}'(x)dx\\ &=\int h(x)dx\\ \end{align}\end{split}\]

\[\int h(x)dx = \int h(g(u))\cdot {g}'(u)du\]

Continue to prove

If \({g}'\) is continuous on the interval \([a,b]\) and \(h\) is continuous on the range of \(g(u,t)=x\), then

\[\int_{g(a)}^{g(b)}h(x)dx=\int_{a}^{b}h(g(u))\cdot {g}'(u)du\]

Proof

Let \(F\) denote any antiderivative of \(h\). Then,

\[\begin{split}\begin{aligned} \int_{a}^{b} h(g(x)) \cdot g^{\prime}(x) d x & =F(g(x))\Bigg ]_{x=a}^{x=b} \\ & =F(g(b))-F(g(a)) \\ & =F(u)\Bigg]_{u=g(a)}^{u=g(b)} \\ & =\int_{g(a)}^{g(b)} h(u) d u . \end{aligned}\end{split}\]