Methods of Integration

(cod: P-70-51-3) Substitution in definite integrals: Let \(u = g(x)\) be a function with a continuous derivative on the interval \([a,b]\) and \(f\) a continuous function on an interval \(J \supset g([a,b])\). We have that \begin{equation*}\label{subst} \int_a^b f(g(x))g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du. \end{equation*} Indeed, let \(F(u)\) be a primitive for the function \(f(u)\) on the interval \(J\). By the Fundamental Theorem of Calculus, we have that \[\int_{g(a)}^{g(b)} f(u) \, du = F(g(b)) - F(g(a)).\] On the other hand, by the Chain Rule, we have that \begin{eqnarray*}\dfrac{d}{dx}F(g(x)) &=& F'(g(x))g'(x) \\ & & \\ &=& f(g(x))g'(x).\end{eqnarray*} Thus, again by the Fundamental Theorem of Calculus, we have that \[\int_a^b f(g(x))g'(x) \, dx = F(g(b)) - F(g(a)),\] which proves the statement.

Often, when performing the substitution \(u = g(x)\) in the context above, we want to replace the integral on the left-hand side with the integral on the right-hand side in the first equation. In other cases, starting from an integral \[\int_A^B f(x) \, dx\] we choose to make a variable change \begin{eqnarray*} x &=& h(\theta), \theta \in I, \\ dx &=& h'(\theta) \, d\theta, \end{eqnarray*} where we take \(h\) injective (invertible) on the interval \(I\), with \(A, B \in h(I)\). By the same argument used above, we then obtain that \[\int_A^B f(x) \, dx = \int_{a}^{b} f(h(\theta)) h'(\theta) \, d\theta,\] where \begin{eqnarray*} A = h(a) &\iff& a = h^{-1}(A) \\ & & \\ B = h(b) &\iff& b = h^{-1}(B), \end{eqnarray*} allowing us to replace the integral on the left-hand side with the one on the right-hand side. The difference in this case compared to the previous one is that, while previously the introduced variable was written as a function of the original variable (\(u = g(x)\)), here we have the opposite, that is, we write the original variable as a function of the new integration variable (\(x = h(\theta)\)).

In the questions below, consider that \(f\) is a continuous function:

(1) By making the substitution \(u = x^2 + 1\), we can verify that the integral \[\int_0^1 2x f(x^2+1) \, dx\] is equal to:

(2) By making the substitution \(x = \sin\, \theta, \theta \in [-\pi/2,\pi/2]\), we can verify that the integral \[\int_0^{1/2} f(x) \, dx\] is equal to:

(1) \(\displaystyle \int_0^1 f(u) \, du\).

(2) \(\displaystyle \int_0^{1/2} f(\sin\, \theta) \cos(\theta) \, d\theta\).

(1) \(\displaystyle \int_1^2 f(u) \, du\).

(2) \(\displaystyle \int_0^{\pi/3} f(\sin\, \theta) \, d\theta\).

(1) \(\displaystyle \int_0^2 f(u) \, du\).

(2) \(\displaystyle \int_0^{\pi/3} f(\sin\, \theta) \cos(\theta) \, d\theta\).

(1) \(\displaystyle \int_1^2 f(u) \, du\).

(2) \(\displaystyle \int_0^{\pi/6} f(\sin\, \theta) \cos(\theta) \, d\theta\).

(1) \(\displaystyle \int_1^2 2 f(u) \, du\).

(2) \(\displaystyle \int_0^{1/2} f(\sin\, \theta) \, d\theta\).