Transform integral from a to b f(x) dx using a linear

Transform $\int_a^b f(x) \, dx$ using a linear substitution into an integral with given limits $\alpha$ and $\beta$.

We want to find a linear transformation that maps the interval $[a, b]$ to the interval $[\alpha, \beta]$.

Step 1: Set up the linear transformation

We seek a linear function $x = g(t) = At + B$, where $A$ and $B$ are constants to be determined, such that when $t=\alpha$, $x=a$, and when $t=\beta$, $x=b$.

Step 2: Determine the constants

From the conditions: $a=A\alpha+B$ (1), $b=A\beta+B$ (2). Subtracting (1) from (2):

\[b - a = A(\beta - \alpha) \quad\Rightarrow\quad A = \frac{b - a}{\beta - \alpha}\]

From (1): $B = a - A\alpha = \frac{a\beta - b\alpha}{\beta - \alpha}$.

Step 3: Write the transformation

\[x = \frac{b - a}{\beta - \alpha} t + \frac{a\beta - b\alpha}{\beta - \alpha} = a + \frac{(b - a)(t - \alpha)}{\beta - \alpha}\]

Step 4: Find $dx$

\[dx = \frac{b - a}{\beta - \alpha} \, dt\]

Step 5: Transform the integral

\[\boxed{\int_a^b f(x) \, dx = \frac{b - a}{\beta - \alpha} \int_\alpha^\beta f\left(a + \frac{(b-a)(t-\alpha)}{\beta-\alpha}\right) dt}\]

Special cases:

Case 1: Transform to $[0, 1]$ ($\alpha = 0$, $\beta = 1$): $x = a + (b - a)t$, $dx = (b - a) \, dt$:

\[\int_a^b f(x) \, dx = (b - a) \int_0^1 f(a + (b-a)t) \, dt\]

Case 2: Transform to $[-1, 1]$ ($\alpha = -1$, $\beta = 1$): $x = \frac{a+b}{2} + \frac{b-a}{2}t$, $dx = \frac{b-a}{2} \, dt$:

\[\int_a^b f(x) \, dx = \frac{b - a}{2} \int_{-1}^1 f\left(\frac{a + b}{2} + \frac{b - a}{2}t\right) dt\]

Verification:

At $t=\alpha$: $x=a+0=a$. At $t=\beta$: $x=a+(b-a)=b$.

Geometric interpretation:

The linear transformation $x = a + \frac{(b-a)(t-\alpha)}{\beta-\alpha}$ performs an affine mapping that scales the interval $[\alpha, \beta]$ by factor $\frac{b-a}{\beta-\alpha}$, shifts and adjusts so that $\alpha \mapsto a$ and $\beta \mapsto b$, and preserves the ratio of subdivisions within the interval.

Final answer:

\[\boxed{x = a + \frac{(b-a)(t-\alpha)}{\beta-\alpha}, \quad dx = \frac{b-a}{\beta-\alpha} \, dt}\]