On which derivative rule is integration by parts based?
Read moreTable of Contents
Textbook Solutions for Calculus: Early Transcendentals
Question
Looking ahead (to Chapter 9) Suppose that a function has derivatives of all orders near r = 0. By the Fundamental Theorem of Calculus,
\(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\)
a. Evaluate the integral using integration by parts to show that
\(f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t\)
b. Show (by observing a pattern or using induction) that by inte
grating by parts n times,
\(f(x)=f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0)+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots\)
This expression, called the Taylor series for f at x = 0, is revisited in Chapter 9.
Solution
The first step in solving 7.1 problem number trying to solve the problem we have to refer to the textbook question: Looking ahead (to Chapter 9) Suppose that a function has derivatives of all orders near r = 0. By the Fundamental Theorem of Calculus,\(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\)a. Evaluate the integral using integration by parts to show that\(f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t\)b. Show (by observing a pattern or using induction) that by integrating by parts n times,\(f(x)=f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0)+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots\)This expression, called the Taylor series for f at x = 0, is revisited in Chapter 9.
From the textbook chapter Integration by Parts you will find a few key concepts needed to solve this.
Visible to paid subscribers only
Step 3 of 7)Visible to paid subscribers only
full solution