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91Ó°ÊÓ

Prove that the Net Change Theorem (Theorem 4.27) is equivalent to the Fundamental Theorem of Calculus.

Short Answer

Expert verified

Ans:∫abf(x)dx=limn→∞∑k=1nFxk-Fxk-1=limn→∞F(b)-F(a)=F(b)-F(a)

Step by step solution

01

Step 1. Given Information: 

The objective is to prove the Fundamental Theorem of Calculus.

Let, fbe a continuous function on [a,b]and F be any anti-derivative of f.

02

Step 2. Proving with the help of the limit of Reimann sum :

So, the definite integral from x=ato x=bdefined by a limit of Reimann sum is,

∫abf(x)dx=limn→∞∑k=1nf(x°k)△x=limn→∞∑k=1nF'(x°k)△x[F'=f]Where,foreachn,△x=b-an,xk=a+k△xandx°kisapointonthesubinterval[xk+1,xk]NOwapplyingtheMeanValueTheoremtoFon[a,b]=[xk+1,xk]suchthatthereexistssomepointsck∈[xk+1,xk]suchthat,F'ck=Fxk-Fxk-1xk-xk-1

03

Step 3. Substituting the above expression for F'(x°k) in the Reimann sum:

∫ab f(x)dx=limn→∞ ∑k=1n F'(xk*)Δx=limn→∞∑k=1n (F(xk)-F(xk-1)(xk-xk-1Δx=(F(x1)-F(x0))+(F(x2)-F(x1))+⋯+(F(x(n-1))-F(x(n-2)))+(F(xn)-F(x(n-1)))=-F(x0)+(F(x1)-F(x1))+(F(x2)-F(x2))+(F(x(n-1))-F(x(n-1)))+F(xn)=-F(x0)+0+0+0+F(xn)=F(b)-F(a)

04

Step 4. Differentiation of the Net change theorem :

TheNetchangetheoremforthefunctionfwhichisdifferentiableon[a,b]is,∫abf'(x)dx=f(b)-f(a)∴∫abf(x)dx=limn→∞∑k=1nFxk-Fxk-1=limn→∞F(b)-F(a)=F(b)-F(a)

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