How to Perform a Scaling Substitution

Step-by-Step Guide

1. 1

   Step 1: Evaluate the integral below.

   There are a few key things that an integral must have in order for scaling substitution to ideally work.
   
   The lower bound must always be 0, and the upper bound must always be a constant, not a variable. ∫05x3(25−x2)dx{\displaystyle \int _{0}^{5}x^{3}(25-x^{2}){\mathrm {d} }x} Nothing is stopping you from expanding the integrand out and evaluating from there, but do you really want to deal with the huge numbers afterwards?
2. 2

   Step 2: Perform a scaling substitution by setting x=5u{\displaystyle x=5u}.

   This does a few things.
   
   First, the upper bound is set to unity.
   
   Second, this works out very nicely, because we see that there is a 25 in the integrand, a number divisible by
   5.
   
   Fives are now being factored out from the integral. ∫0153u3(25−52u2)5du{\displaystyle \int _{0}^{1}5^{3}u^{3}(25-5^{2}u^{2})5{\mathrm {d} }u} We can clearly see that 56{\displaystyle 5^{6}} can be factored out of the integral.
   
   In fact, with practice, we can see this from the very start just by adding the exponents and remembering to add 1 for the dx{\displaystyle {\mathrm {d} }x} as well.
   
   Remember, dx=5du.{\displaystyle {\mathrm {d} }x=5{\mathrm {d} }u.} 56∫01u3(1−u2)du{\displaystyle 5^{6}\int _{0}^{1}u^{3}(1-u^{2}){\mathrm {d} }u} , 56∫01u3(1−u2)du=56(14−16)=56224=56223{\displaystyle {\begin{aligned}5^{6}\int _{0}^{1}u^{3}(1-u^{2}){\mathrm {d} }u&=5^{6}\left({\frac {1}{4}}-{\frac {1}{6}}\right)\\&=5^{6}{\frac {2}{24}}\\&={\frac {5^{6}}{2^{2}3}}\end{aligned}}} Through this evaluation, we dealt with very small and simple numbers.
   
   By integrating normally, we would've dealt with numbers in the thousands and ten-thousands. , We have a nested exponent and a 3{\displaystyle {\sqrt {3}}} in the upper bound now, but that does not dissuade us from scaling. ∫03x6(3−x2)3dx{\displaystyle \int _{0}^{\sqrt {3}}x^{6}(3-x^{2})^{3}{\mathrm {d} }x} , The nested exponent counts for 6, because the quantity inside counts for 2, and that is being cubed.
   
   Adding the factors up, we get 313/2{\displaystyle 3^{13/2}} outside of the integral. 313/2∫01u6(1−u2)3du{\displaystyle 3^{13/2}\int _{0}^{1}u^{6}(1-u^{2})^{3}{\mathrm {d} }u} , Integrals that allow for scaling substitution can often yield multiple fractions with non-common denominators, so your arithmetic skills are put to the test. 313/2∫01u6(1−u2)3du=313/2∫01u6(1−3u2+3u4−u6)du=313/2(17−39+311−113)=313/2(327⋅11−163⋅13)=313/224(27⋅11−13⋅13)=313/224(78−773⋅7⋅11⋅13)=311/2247⋅11⋅13{\displaystyle {\begin{aligned}3^{13/2}\int _{0}^{1}u^{6}(1-u^{2})^{3}{\mathrm {d} }u&=3^{13/2}\int _{0}^{1}u^{6}(1-3u^{2}+3u^{4}-u^{6}){\mathrm {d} }u\\&=3^{13/2}\left({\frac {1}{7}}-{\frac {3}{9}}+{\frac {3}{11}}-{\frac {1}{13}}\right)\\&=3^{13/2}\left({\frac {32}{7\cdot 11}}-{\frac {16}{3\cdot 13}}\right)\\&=3^{13/2}2^{4}\left({\frac {2}{7\cdot 11}}-{\frac {1}{3\cdot 13}}\right)\\&=3^{13/2}2^{4}\left({\frac {78-77}{3\cdot 7\cdot 11\cdot 13}}\right)\\&={\frac {3^{11/2}2^{4}}{7\cdot 11\cdot 13}}\end{aligned}}} , We will show that scaling substitution can still work to an extent when there is not a good factor in the integrand.
   
   In this case, we have a 40 instead of a
   36. ∫06x4(40−x2)(6+x)dx{\displaystyle \int _{0}^{6}x^{4}(40-x^{2})(6+x){\mathrm {d} }x} , ∫06x4(40−x2)(6+x)dx=66∫01u4(40−36u2)(1+u)du=6622∫01u4(10−9u2)(1+u)du{\displaystyle {\begin{aligned}\int _{0}^{6}x^{4}(40-x^{2})(6+x){\mathrm {d} }x&=6^{6}\int _{0}^{1}u^{4}(40-36u^{2})(1+u){\mathrm {d} }u\\&=6^{6}2^{2}\int _{0}^{1}u^{4}(10-9u^{2})(1+u){\mathrm {d} }u\end{aligned}}} , Although we were not able to factor out a 62,{\displaystyle 6^{2},} we were still able to get a 22{\displaystyle 2^{2}} out, which is a simplification.
   
   Note that 211 is a prime number. 6622∫01u4(10−9u2+10u−9u3)du=6622(105−97+106−98)=6622(21−97+53−98)=6622(113−1357⋅8)=6622(616−4057⋅3⋅23)=3525⋅2117{\displaystyle {\begin{aligned}6^{6}2^{2}\int _{0}^{1}u^{4}(10-9u^{2}+10u-9u^{3}){\mathrm {d} }u&=6^{6}2^{2}\left({\frac {10}{5}}-{\frac {9}{7}}+{\frac {10}{6}}-{\frac {9}{8}}\right)\\&=6^{6}2^{2}\left({\frac {2}{1}}-{\frac {9}{7}}+{\frac {5}{3}}-{\frac {9}{8}}\right)\\&=6^{6}2^{2}\left({\frac {11}{3}}-{\frac {135}{7\cdot 8}}\right)\\&=6^{6}2^{2}\left({\frac {616-405}{7\cdot 3\cdot 2^{3}}}\right)\\&={\frac {3^{5}2^{5}\cdot 211}{7}}\end{aligned}}} If this evaluation seems scary to you, consider that the numerator of the answer is greater than one million.
3. 3

   Step 3: Evaluate by any means possible.

4. 4

   Step 4: Evaluate the integral below.

5. 5

   Step 5: Set x=3u{\displaystyle x={\sqrt {3}}u}.

6. 6

   Step 6: Evaluate.

7. 7

   Step 7: Evaluate the integral below.

8. 8

   Step 8: Set x=6u{\displaystyle x=6u}.

9. 9

   Step 9: Evaluate.

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