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In chapter 4, section 5 of the 7th edition of James Stewart's Calculus book is the following problem.

51.) Evaluate $\displaystyle{\int_{0}^{1} \frac{1}{(1 + \sqrt{x})^4} \, dx}$.

In this section, only the Substitution Rule was presented.

Substitution Rule for Definite Integrals

If $g^{\prime}$ is continuous on the interval $[a, b]$ and $f$ is continuous on the image under $g$ of $[a, b]$, \begin{equation} \int_{a}^{b} f\bigl(g(x)\bigr) g^{\prime}(x) \, dx = \int_{g(a)}^{g(b)} f(x) \, dx . \end{equation}

What solution is intended by Stewart using this rule?

I understand that with the substitution $x = t^{2}$, the definite integral can be evaluated using Integration by Parts. This is not presented until Chapter 7 in this textbook.

This problem is not in the section on the Substitution Rule in the previous edition of the textbook. I suspect that the problem was misplaced in the 7th edition.

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    $\begingroup$ What would be a $g$ such that $f(g(t))$ is nicer than the given integrand? $\endgroup$ Feb 24, 2016 at 21:35
  • $\begingroup$ You could do a substitution with $u=\sqrt{x}$, but that requires partial fractions. $\endgroup$
    – zz20s
    Feb 24, 2016 at 21:35
  • $\begingroup$ @zz20s Yep. Partial Fractions is not discussed until Chapter 7. $\endgroup$
    – user232552
    Feb 24, 2016 at 21:37
  • $\begingroup$ If you let u = 1 + sqrt(x) <==> sqrt(x)=(u-1), then du = dx/(2*sqrt(x)), so dx = 2*sqrt(x)*du = 2*(u-1)*du, and then it's easy. $\endgroup$
    – Name
    Feb 24, 2016 at 21:52
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    $\begingroup$ Another question about the same integral was posted recently: Definite Integral: $\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$. $\endgroup$ Jan 27, 2018 at 12:29

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Let $u=1+\sqrt{x}$, so $x=(u-1)^2$ and $dx=2(u-1)du$.

Then $\displaystyle\int_0^1\frac{1}{(1+\sqrt{x})^4}dx=\int_1^2\frac{1}{u^4}\cdot2(u-1)du=2\int_1^2(u^{-3}-u^{-4})du=2\left[-\frac{1}{2}u^{-2}+\frac{1}{3}u^{-3}\right]_1^2$

$\displaystyle=2\left(-\frac{1}{8}+\frac{1}{24}-\big(-\frac{1}{2}+\frac{1}{3}\big)\right)=2\left(\frac{1}{12}\right)=\frac{1}{6}$


The Substitution Rule as stated does not apply to this example, since $g^{\prime}(x)=\frac{1}{2\sqrt{x}}$ is not continuous on $[0,1]$.

However, if $f(u)=2(u^{-3}-u^{-4}),\;\;$ then $\displaystyle f(g(x))g^{\prime}(x)=\frac{1}{(1+\sqrt{x})^4}$ for $x>0$;

so it is continuous on $[0,1]$ if we define its value at 0 to be 1,

and therefore the conclusion of the Substitution Rule still applies.

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  • $\begingroup$ I think the limits on the third integral should be 1 to 2. $\endgroup$ Feb 24, 2016 at 23:30
  • $\begingroup$ @martycohen Thanks - I'll correct this. $\endgroup$
    – user84413
    Feb 25, 2016 at 0:25
  • $\begingroup$ @user84413 This is correct. I alluded to an analogous argument in my post. I am asking whether there is a solution that uses the Substitution Rule as stated in his textbook (and repeated in my post)? $\endgroup$
    – user232552
    Feb 25, 2016 at 2:09
  • $\begingroup$ I think you are making a valid point; the Substitution Rule as stated does not apply to this example. I will edit my answer to address this, although some MSE users might have a better response to this point. $\endgroup$
    – user84413
    Feb 25, 2016 at 17:34
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Replace $x$ with $u^2$ to get $dx=2u\,du$ then: $$ I = 2\int_{0}^{1}\frac{u}{(1+u)^4}\,du = 2\int_{0}^{1}\frac{du}{(1+u)^3}-2\int_{0}^{1}\frac{du}{(1+u)^4}, $$ from which: $$ I = 2\cdot\left[-\frac{1}{2}\cdot\frac{1}{(1+u)^2}+\frac{1}{3}\cdot\frac{1}{(1+u)^3}\right]_{0}^{1} = 2\cdot\frac{1}{12}=\color{red}{\frac{1}{6}}.$$

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  • $\begingroup$ This is a good solution, but doesn't it use partial fractions? $\endgroup$
    – zz20s
    Feb 24, 2016 at 21:39
  • $\begingroup$ @Jack D'Aurizio Maybe I should have given this solution in my post. Such a technique of integration is not presented in this textbook until Chapter 7. $\endgroup$
    – user232552
    Feb 24, 2016 at 21:41
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    $\begingroup$ It does not use partial fractions. It uses $u = (1+u)-1$, cancels one $1+u$, and then the integral of $1/(1+u)^n$ for $n=3, 4$. $\endgroup$ Feb 24, 2016 at 23:32
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I think this may be what you mean, without partial fractions, integration by parts or other more advanced methods:

Substitute $\;u=1+\sqrt x\implies du=\frac{dx}{2\sqrt x}\implies dx=2(u-1)\,du\;$ , and

$$\int_0^1\frac{dx}{(1+\sqrt x)^4}=\int_1^2\frac{2(u-1)\,du}{u^4}=2\int_1^2\frac{du}{u^3}-2\int_1^2\frac{du}{u^4}=$$

$$=\left.-\frac1{u^2}\right|_1^2+\left.\frac23\frac1{u^3}\right|_1^2=-\frac14+1+\frac2{24}-\frac23=\frac16$$

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    $\begingroup$ Using the Substitution Rule, as stated in the textbook, you are saying to define $g(x) = 1 + \sqrt{x}$. Its derivative is $g^{\prime}(x) = (1/2)x^{-1/2}$. $g^{\prime}$ is not continuous on the interval $[0, 1]$. So, the Substitution Rule cannot be applied using this definition for $g$. $\endgroup$
    – user232552
    Feb 24, 2016 at 22:40
  • $\begingroup$ This is not a reliable textbook. My guess is that you are right - your solution is probably what he intended. $\endgroup$
    – user232552
    Feb 24, 2016 at 22:40
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Try the change of variables $x(t) := t^2$ that maps $[0,1]$ to itself. It is positive, bijective and continuously differentiable, and yields

$$ \int_{0}^1 \frac{\mathrm{d}x}{(1+\sqrt x)^4} = \int_0^1 \frac{2t}{(1+t)^4}\mathrm d t $$ which can be readily integrated by parts, or computed straightaway by noticing that $t = 1 + t -1$.

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  • $\begingroup$ Everybody is replying with solutions like this. Such a technique of integration is not presented in this textbook until Chapter 7. $\endgroup$
    – user232552
    Feb 24, 2016 at 21:42
  • $\begingroup$ Everyone seems to be suggesting that. Just integrating the $1/(1+t)^4$ denominator and differentiating $2t$ by integration by parts does just fine, though. Or is integration by parts presented later ? $\endgroup$
    – ManifoldFR
    Feb 24, 2016 at 21:44
  • $\begingroup$ I suppose the question is: can @Adelyn use integration by parts? $\endgroup$
    – zz20s
    Feb 24, 2016 at 21:45
  • $\begingroup$ @zz20s 21 I will edit my post. Integration by Parts is presented in Chapter 7. This could not be the intended solution ... unless the problem was misplaced. $\endgroup$
    – user232552
    Feb 24, 2016 at 21:46
  • $\begingroup$ Alternatively, you could always write $t = 1 + t - 1$ on the right hand side, which is essentially what @Nehorai did. $\endgroup$
    – ManifoldFR
    Feb 24, 2016 at 21:52

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