BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 6.5, Problem 21E
To determine

To prove: The mean value theorem for the integral function F(x)=axf(t)dt.

Expert Solution

Answer to Problem 21E

The mean value theorem of integrals is f(c)=1baabf(x)dx.

Explanation of Solution

Show the part 2 of fundamental theorem of calculus.

F(x)=axf(t)dt

Given information:

The function is ddx[axf(t)dt]=f(x).

Show the expression for mean value theorem of integrals.

f(c)=favg=1baabf(x)dx.

Applying the condition for mean value theorem for integrals.

  • If f is continuous on closed interval [a,b].
  • If F is differentiable on open interval (a,b).

There is a exists number of c. Hence, the expression of function as follows:

f(c)=f(b)f(a)ba.

Consider F(x)=axf(t)dt.

Apply mean value theorem for derivative.

F(c)=F(b)F(a)baddx(F(c))=F(b)F(a)ba (1)

Replace [axf(t)dt]c for F(c) in Equation (1).

ddx[axf(t)dt]c=F(b)F(a)ba

Apply part 2 of fundamental theorem of calculus.

f(c)=1ba[(abf(x)dxabf(x)dx)]f(c)=1baabf(x)dx

Thus, The mean value theorem of integrals is f(c)=1baabf(x)dx.

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