How to change integral bounds from [a,b] to [c,d]?

Question

How can I change those bounds - using assigmnets of t=...: $$\int_{0.5}^{1.5} e^{x^2} dx$$
To those bounds: $$0.5\int_{-1}^{1} e^{-(1+0.5t)^2} dt$$

I've seen that if $x=1+0.5t$ it will work, but I don't know how to do it in general. For example if next time my bounds were [a,b] and I needed the bounds to be [c,d] with some assignment

Answer 1

In general, you have the range $x = a$ to $x = b$, and you want to find $t = Px + Q$ so that $t$ will go from $c$ to $d$. In other words, you want

\begin{align} Pa + Q &= c \\ Pb + Q &= d \\ \end{align} and you need to find $P$ and $Q$. Then you'll do the substitution $$ t = Px + Q \\ dt = P ~ dx\\ \frac{1}{P} ~dt = dx $$

So how do you find $P$ and $Q$? Remember that $a, b, c, d$ are known. So write \begin{align} Pa + Q &= c \\ Pb + Q &= d \\ \end{align} Subtract to get: \begin{align} P(a-b) &= c-d \\ \end{align} Solve for $P$: \begin{align} P &= \frac{c-d}{a-b} \end{align} And now take that (known) value of $P$ and use it in $$ Q = c - Pa = c - \left( a \frac{c-d}{a-b} \right) $$ to find $Q$.

And that's all you need to do!

Answer 2

Substitute $x=1+\frac t 2$.... Then $dx=\frac {dt} 2$

And $-1 \le t \le 1$

$$\int_{\frac 1 2}^{ \frac 3 2}e^{x^2}dx=\frac 1 2 \int_{-1}^1 e^{(1+\frac t 2)^2}dt$$

Try to find the values of a,b in $x=at+b$

Since you know that:

$$ \begin{cases} \frac 1 2=a(-1)+b \\ \frac 3 2=a(1)+b \end{cases} $$

Answer 3

let's say you want to compute $$\int_a^b f(x)dx$$

I understand that you want to transform the above integral into something of the form $$\int_c^d f(g(u))d(g(u))$$

using some change of variable $x = g(u)$

now how to choose $g$ ?

very simple

you just have to have $a = g(c)$ and $b = g(b)$

to simplify the computations choose $g$ to be a first order polynomial

$$g(x) = a_1x + a_0$$

so it only remains to determine $a_1$ and $a_2$

which is very simple because it only suffices to solve the following $2 \times 2$ system

$\begin{align} a = a_1c + a_0 \\ b = a_1d + a_0 \\ \end{align}$

where $a_1$ and $a_0$ are the unknowns

Answer 4

You want to change $$\int_a^b e^{x^2}\,dx\mapsto\int_c^d e^{x^2}\,dx$$ We need $x\mapsto f(x)$ where $f(a)=c$, $f(b)=d$. Assuming $f$ is linear, $$f(x)=mx+k\\c=ma+k\\d=mb+k\\m=\frac{d-c}{b-a}\\k=d-\frac{d-c}{b-a}\cdot b=\frac{db-da-db+cb}{b-a}=\frac{bc-ad}{b-a}$$

So the change of variables you are after is $$t=f(x)=\frac{(d-c)x+bc-ad}{b-a}$$

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In the example you gave, $f(x)$ simplifies to $t=2x-2$, or equivalently, $x=\frac{t+2}2=1+\frac t2$, as required.

Answer 5

If your bounds are $[a,b]$ and you wish that your bounds were $[c,d]$, you can always make a linear change of variables as follows:

1. If your bounds of integration are $[a,b]$ and you change variables $t(x) = \frac{x-a}{b-a}$, then the bounds of integration change to $[0,1]$. After all, you can confirm $t(a) = 0$ and $t(b)=1$.

2. If your bounds of integration are $[0,1]$ and you change variables $u(t) = (d-c)t + c$, then the bounds of integration change from $[0,1]$ to $[c,d]$. After all, $u(0) = c$ and $u(1)=d$.

3. Applying both of these changes, if your bounds of integration are $[a,b]$ and you wish they were $[c,d]$, you can change bounds using the formula: $$y(x) = \frac{d-c}{b-a}(x-a) + c$$ This formula works in general, regardless of whether $a>b$ and so on.

4. In this specific case, your bounds are $[a,b] = [\frac{1}{2},\frac{3}{2}]$ and you wish they were $[c,d] = [-1, +1]$. According to this rule, the transformation you need is:

$$t(x) = \frac{d-c}{b-a}(x-a) + c = \frac{2}{1}(x-\frac{1}{2}) + (-1) = 2(x-1)$$