Question

Use integration by substitution to show that if \(y\) is a continuous function of \(x\) on the interval \(a \leq x \leq b\), where \(x=f(t)\) and \(y=g(t),\) then $$ \int_{a}^{b} y d x=\int_{t_{1}}^{t_{2}} g(t) f^{\prime}(t) d t $$ where \(f\left(t_{1}\right)=a, f\left(t_{2}\right)=b,\) and both \(g\) and \(f^{\prime}\) are continuous on \(\left[t_{1}, t_{2}\right]\)

Short answer

By using integration by substitution, we can show that \[ \int_{a}^{b} y dx = \int_{t_1}^{t_2} g(t) f^\prime(t) dt \] where the limits of integration change from \([a, b]\) to \([t_1, t_2]\) due to the change of variable.

Step-by-step solution

Write down the given information

We know that \(x = f(t)\), \(y = g(t)\) and \(\int_{a}^{b} y dx\) . Our goal is to express this integral in terms of \(t\) using the function \(f\) and its derivative. Also note that the limits \(a\) and \(b\) correspond to \(t_1\) and \(t_2\), with \(f(t_1)=a\) and \(f(t_2)=b\). Therefore, when \(x = a\), \(t = t_1\) and when \(x = b\), \(t = t_2\).

Perform the substitution

We will substitute \(x = f(t)\) and \(y = g(t)\) into the integral and change the differential \(dx\) to \(dt\) using the chain rule. The chain rule states that \[ dx = f^\prime(t)dt \]. Thus the integral becomes \[ \int_{a}^{b} y dx = \int_{a}^{b} g(t) f^\prime(t) dt \]

Adjust the limits of integration

Lastly, since we changed the variable of integration from \(x\) to \(t\), we also need to adjust the limits of integration accordingly. The lower limit changes from \(a\) to \(t_1\) and the upper limit changes from \(b\) to \(t_2\). Thus, the integral becomes \[ \int_{a}^{b} g(t) f^\prime(t) dt = \int_{t_1}^{t_2} g(t) f^\prime(t) dt \]