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This is the integral $$\int_1^3 \frac{4}{(2x-3)^4} dx $$

Solving by u-substitution, it works fine. $$ \text{let}~~ u = 2x-3 $$

$$\int_1^34 \cdot ( 2x-3 )^{-4}dx $$

$$\ x = \frac{u+3}{2} $$

$$\ \frac{ dx }{ du }=\frac{1 }{2} $$ $$\ dx = \frac{ 1 }{ 2 }du $$

EDIT: $$\ u(3) = 2(3)- 3$$ $$\ u(3) = 3 $$ $$\ u(1) = 2(1)- 3$$ $$\ u(1) = -1 $$ $$\int_{-1}^3 4 \cdot u^-4 \cdot \frac{ 1 }{ 2 }du $$

$$\int_{-1}^3 2\cdot u^{-4}du $$

$$\frac{ 2u^{-3}}{-3 } \Bigg \vert_{-1}^{3} \ $$

$$ = \frac{ 2}{-3(2x-3)^3 } \Bigg \vert_{-1}^{3} \ $$

$$ = \frac{ 2}{ -3(2(3)-3)^3 } - \frac{ 2}{ -3(2(-1)-3)^3} $$ $$ = \frac{-2}{81} - \frac{2}{375} $$ $$ = \frac{-84}{125} $$

But plugging this integral into a TI-84 CE Plus throws an error "Cannot divide by 0"

Also trying online calculators, Wolframalpha and Freemathhelp, the problem is unable to be solved.

Why can't this problem be solved on these calculators?

Oreofe Solarin
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    It's an improper integral (the function is not defined for $$x=3/2$$) and the improper integral does not converge – Nick Castillo Dec 18 '21 at 18:55
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    A simpler example of this wrong reasoning is $\int_{-1}^1 \frac 1{x^2}dx = (-\frac 1x) |_{-1}^1 = -2$. – Note also that the result cannot be correct because the integral over a positive function can not be negative. – Martin R Dec 18 '21 at 19:00
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    I don't understand why this has downvotes. It has a clear attempt. – Shaun Dec 18 '21 at 19:23
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    Note that you made in mistake in the substitution, namely the bounds are wrong. The new interval is $[-1,3]$ not $[1,3]$. And $\dfrac 2{u^4}$ is not defined in $0\in[-1,3]$ and your integral is still improper after substitution. – zwim Dec 18 '21 at 23:28
  • @zwim Why does the interval change to $$\ [-1,3] $$ – Oreofe Solarin Dec 19 '21 at 05:45
  • The interval changes due to your substitution. @Oreofe – Laxmi Narayan Bhandari Dec 19 '21 at 11:45
  • You have $u(x)=2x-3$ so the bounds $u(1)=-1$ and $u(3)=3$. When you do u-subs, the bounds must be changed to $\displaystyle\int_{u(1)}^{u(3)} f(u)du$ – zwim Dec 19 '21 at 14:22
  • @zwim the interval change still does explain why the integral cannot be solved on calculators. – Oreofe Solarin Dec 19 '21 at 18:38
  • The integral doesn't make sense because of the singularity at $x = 3/2.$ Sure, you can use a u-sub, but does the integral make sense? No. That's why calculators can't do it because they're approximating something that looks like, say, $\frac{1}{10^{-32}}.$ – Sean Roberson Dec 19 '21 at 18:42
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    Also, ask yourself - how does a positive function have negative area? – Sean Roberson Dec 19 '21 at 19:24
  • At first I thought your problem was using an inferior TI calculator. My HP Prime returned infinity on this one. Running the numerical integrator on my WP-34s did not converge; last time I looked it was around 6,201,000,000, which is mathematically defined as "almost infinity". – richard1941 Dec 21 '21 at 19:14

2 Answers2

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When we reason$$\int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a),$$we assume every $g$ with $g^\prime=f$ satisfies $g(b)-g(a)=F(b)-F(a)$, so the choice of $f$'s antiderivative is irrelevant.

But in general, antiderivatives differ additively by locally constant functions on the integrand's domain, which due the discontinuity at $x=\frac32$ is in this case in two components as $[1,\,3]\setminus\left\{\frac32\right\}=\left[1,\,\frac32\right)\cup\left(\frac32,\,4\right]$. We can make this even more technical, but that's all we need for now.

A locally constant function on this domain can have different values either side of $\frac32$, say $C_-$ on the left and $C_+$ on the right. But if we add such a function to an antiderivative of $4(2x-3)^{-4}$, the difference between the antiderivatives' values at $a,\,b$ with $a<\frac32<b$ changes by $C_+-C_-$, which is in general nonzero, and so the integral is improper.

J.G.
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OP: the interval change still does explain why the integral cannot be solved on calculators

This is because the "dysfunction"/discrepancy that you are raising has nothing to do with the integration-by-substitution process.

If $f$ is Riemann integrable on $[a,b]$ and has antiderivative $F,$ the second fundamental theorem of calculus guarantees that $$\int_a^b f=F(b)-F(a).$$ In your example, the boldfaced condition is not satisfied (the integral fails to converge/exist due to the singularity at $x=1.5$), and $$\int_1^3 \frac{4}{(2x-3)^4} \mathrm dx\ne \left[\frac2{3 (3 - 2 x)^3}\right]_1^3,$$ even though $$\int\frac{4}{(2x-3)^4} \mathrm dx=\frac2{3 (3 - 2 x)^3}+C.$$

This, together with Martin's example in the second comment, shows that an integrand may have an antiderivative whilst not being integrable on the desired interval.

J.G.
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ryang
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