{"id":405,"date":"2025-02-13T19:44:39","date_gmt":"2025-02-13T19:44:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-using-substitution-learn-it-2\/"},"modified":"2025-02-13T19:44:39","modified_gmt":"2025-02-13T19:44:39","slug":"integration-using-substitution-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-using-substitution-learn-it-2\/","title":{"raw":"Integration using Substitution: Learn It 2","rendered":"Integration using Substitution: Learn It 2"},"content":{"raw":"\n

Substitution for Definite Integrals<\/h2>\n

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.<\/p>\n

\n
\n

substitution with definite integrals<\/h3>\n

Let [latex]u=g(x)[\/latex] and let [latex]{g}^{\\text{\u2032}}[\/latex] be continuous over an interval [latex]\\left[a,b\\right],[\/latex] and let [latex]f[\/latex] be continuous over the range of [latex]u=g(x).[\/latex] Then,<\/p>\n

[latex]{\\displaystyle\\int }_{a}^{b}f(g(x)){g}^{\\prime }(x)dx={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du[\/latex]<\/div>\n<\/div>\n<\/section>\n

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if [latex]F(x)[\/latex] is an antiderivative of [latex]f(x),[\/latex] we have<\/p>\n

[latex]\\displaystyle\\int f(g(x)){g}^{\\prime }(x)dx=F(g(x))+C[\/latex]<\/div>\n

Then<\/p>\n

[latex]\\begin{array}{cc}{\\displaystyle\\int }_{a}^{b}f\\left[g(x)\\right]{g}^{\\prime }(x)dx\\hfill & ={F(g(x))|}_{x=a}^{x=b}\\hfill \\\\ & =F(g(b))-F(g(a))\\hfill \\\\ & ={F(u)|}_{u=g(a)}^{u=g(b)}\\hfill \\\\ \\\\ \\\\ & ={\\displaystyle\\int }_{g(a)}^{g(b)}f(u)du,\\hfill \\end{array}[\/latex]<\/div>\n

and we have the desired result.<\/p>\n

\n

Use substitution to evaluate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[\/latex]<\/p>\n

[reveal-answer q=\"fs-id1170572587723\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572587723\"]<\/p>\n

Let [latex]u=1+2{x}^{3},[\/latex] so [latex]du=6{x}^{2}dx.[\/latex]<\/p>\n

Since the original function includes one factor of [latex]x^2[\/latex] and [latex]du=6{x}^{2}dx,[\/latex] multiply both sides of the du<\/em> equation by [latex]\\frac{1}{6}.[\/latex]<\/p>\n

Then,<\/p>\n

[latex]\\begin{array}{ccc}du\\hfill & =\\hfill & 6{x}^{2}dx\\hfill \\\\ \\frac{1}{6}du\\hfill & =\\hfill & {x}^{2}dx.\\hfill \\end{array}[\/latex]<\/div>\n

To adjust the limits of integration, note that when [latex]x=0,u=1+2(0)=1,[\/latex] and when [latex]x=1,u=1+2(1)=3.[\/latex]<\/p>\n

Then,<\/p>\n

[latex]{\\displaystyle\\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du.[\/latex]<\/div>\n

Evaluating this expression, we get<\/p>\n

[latex]\\begin{array}{}\\frac{1}{6}{\\displaystyle\\int }_{1}^{3}{u}^{5}du\\hfill & =(\\frac{1}{6})(\\frac{{u}^{6}}{6}){|}_{1}^{3}\\hfill \\\\ & =\\frac{1}{36}\\left[{(3)}^{6}-{(1)}^{6}\\right]\\hfill \\\\ & =\\frac{182}{9}.\\hfill \\end{array}[\/latex]<\/div>\n

Watch the following video to see the worked solution to this example.<\/p>\n