   Chapter 7.1, Problem 70E

Chapter
Section
Textbook Problem

# If f ( 0 ) = g ( 0 ) = 0 and f ′ ′ and g ′ ′ are continuous, show that ∫ 0 a f ( x ) g ′ ′ ( x ) d x = f ( a ) g ′ ( a ) − f ′ ( a ) g ( a ) + ∫ 0 a f ′ ′ ( x ) g ( x ) d x

To determine

To show: 0af(x)g(x)dx=f(a)g(a)f(a)g(a)+0af(x)g(x)dx

Explanation

The technique of integration by parts comes in handy when the integrand involves product of two functions. It can be thought of as a rule corresponding to the product rule in differentiation.

Formula used:

The formula for integration by parts for definite integral is given by

abf(x)g(x)dx=f(x)g(x)]ababg(x)f(x)dx

Given:

Values of the functions, f(0)=g(0)=0

Functions f and g are continuous

Calculation:

Consider the left-hand side of the equation:

0af(x)g(x)dx

Solve the integration using integration by parts. Make the choice for u and dv such that the resulting integration from the formula above is easier to integrate. Let

u=f(x)      dv=g(x)dx

Then, the differentiation of u and anti derivative of dv will be

du=f(x)dx      v=g(x)

Substitute for variables in the formula above to get

0af(x)g(x)dx=f(x)g(x)]0a0ag(x)f(x)dx …… (1)

Solve the integral in the last term by again applying integration by parts with variables as:

u=f(x)      dv=g(x)dx

Then, the differentiation of u and antiderivative of dv will be

du=f(x)dx      v=g(x)

Then the integral 0ag(x)f(x)dx will be:

0ag<

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