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Why can't we substitute $(x-2)^2=t$ in the Definite Integral $$\int_0^3(x-2)^2dx$$? And what about the same Substitution in the indefinite integral $$\int(x-2)^2dx$$

user679770
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  • "Why can't we substitute ..." Have you tried that substitution on your own? –  Sep 29 '19 at 16:43
  • You can, but it will make the integral more complicated. See @HansLundmark 's comment – Andrei Sep 29 '19 at 16:43

3 Answers3

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You can definitely substitute $(x-2)^2$ for $t$ in the integral. However, you would also have to to change $dx$ into $dt$ to able to integrate it. Hence substitution has to be done wisely in order to bring the integral into a simpler standard form. It is the same for definite and indefinite.

Also know that in definite integral, the limits change when $dx$ is replaced.

For the proposed problem, the easiest way to integrate would be to simplify $(x-2)^2$ to $x^2-4x+4$ and then its simple.

Sam
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  • But if we substitute $(x-2)^2=t$ then we are using the Substitution of the form $t=g(x)$ and in this $g(x)$ must lie within the limits $g(a)$ and $g(b)$ but this is not the case in the above question as $g(x)=(x-2)^2$ does not lie within the limits $g(0)=4$ and $g(1}=1$ – user679770 Sep 29 '19 at 17:00
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In definite integrals in the domain $(a,b)$ of integration a substitution $t=g(x)$ needs to be continuous and monotonic (increasing or decreasing with out max or min). Otherwise, the subitution is bad one. Interestingly, a bad substitution can be made to work by reducing/breaking the domain of integration. For instance, $t=\tan x$ is a bad substitution for $(0, \pi)$ but it works well for $(0,\pi/2) \cup (\pi/2,\pi)$.

In this question $t=(x-2)^2$ has minimum at $x=2$ in the domain $x\in (0,3)$ so let us break the domain as

$$I=\int_{0}^{2} (x-2)^2 dx+ \int_{2}^{3} (x-2)^2 dx$$ Next, now we can use $$t=(x-2)^2 \Rightarrow x= 2 \mp \sqrt{t} \Rightarrow \mp \frac{1}{2} t^{-1/2}$$ as $$I=\int_{4}^{0} t~ (-\frac{1}{2}t^{-1/2} )~ dt+ \int_{0}^{1} t ~(+\frac{1}{2} t^{-1/2})~dt= \frac{8}{3}+\frac{1}{3}=3.$$ Which is the same as obtainedotherwise directly without a subtitution $$I=\int_{0}^{3} (x-2)^2 dx=\frac{(x-2)^3}{3}|_{0}^{3}=3.$$

Z Ahmed
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  • But how can we substitute $t=$tan$x$ as tan$x$ is discontinuous at $x=\frac{π}{2}$ – user679770 Sep 29 '19 at 17:22
  • @user679770 See when you are integrating $\int_{0}^{\pi/2} f(x) dx$ the domain is $(0,\pi/2)$ it is never $ [0,\pi/2]$ . Similarly, when you are doing $\int_{\pi/2}^{\pi} f(x) dx$ the domain is $(\pi/2, \pi)$ it is never $[\pi/2, \pi]..$ But when your are doing $\int_{0}^{\pi} f(x) dx $ and using $t=\tan x$ the discontinuity is encountered at $x=\pi$ which is very much inside the domain $(0,\pi)$. – Z Ahmed Sep 29 '19 at 17:34
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(1) The reason is that the function $(x-2)^2\mapsto t$ defined on $[0,3]$ by $(x-2)^2=t$ is not monotonic. Thus, it has no well-defined inverse.

(2) You can make this substitution if you want to use the indefinite integral to compute a definite integral on an interval where the above function is monotonic.

Allawonder
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