Mth 3B: Lecture 9 Noh White October 18, 2017
The definite integrl Defintion The definite integrl of function f (x) is defined to be where x = b n. f (x) dx = lim n x n f ( + k x) k=1
Properties of definite integrls Zero re f (x) dx = 0
Properties of definite integrls Adding res c f (x) dx = c f (x) dx + f (x) dx b
More properties of definite integrls Reversing the re f (x) dx = b f (x) dx
More properties of definite integrls Reversing the re f (x) dx = b f (x) dx Additivity f (x) + g(x) dx = f (x) dx + g(x) dx
More properties of definite integrls Reversing the re f (x) dx = b f (x) dx Additivity f (x) + g(x) dx = f (x) dx + g(x) dx Linerrity (sclrs fctor out) αf (x) dx = α f (x) dx
The fundmentl theorem of clculus Theorem For ny, d x f (t) dt = f (x) dx
The fundmentl theorem of clculus Theorem For ny, d x f (t) dt = f (x) dx
The fundmentl theorem of clculus Theorem For ny, Tht is, F (x) = x Note d x f (t) dt = f (x) dx f (t) dt is n ntiderivtive of f (x)!
The fundmentl theorem of clculus Theorem For ny, Tht is, F (x) = x Note F (x) = x d x f (t) dt = f (x) dx f (t) dt is n ntiderivtive of f (x)! f (t) dt is function of x.
The fundmentl theorem of clculus Theorem For ny, Tht is, F (x) = x Note d x f (t) dt = f (x) dx f (t) dt is n ntiderivtive of f (x)! F (x) = x f (t) dt is function of x. every input x produces number s n output.
A consequence (corrollry) Corollry For ny ntiderivtive F (x) of f (x) f (x) dx = F (b) F ()
A consequence (corrollry) Corollry For ny ntiderivtive F (x) of f (x) f (x) dx = F (b) F () Why? Well F (x) = x f (t) dt + C for some nd C. So F (b) F () = = f (t) dt + C f (t) dt f (t) dt C
Exmple 1 Question Evlute the definite integrl 1 0 x 2 4 dx
Exmple 1 Question Evlute the definite integrl 1 0 x 2 4 dx Solution An ntiderivtive of x 2 4 is 1 3 x 3 4x so 1 0 x 2 4 dx = 1 3 13 4 1 3 03 + 4 0 = 1 3 4 = 11 3
Exmple 2 Question Evlute the definite integrl π sin x dx 0
Exmple 2 Question Evlute the definite integrl π 0 sin x dx Solution An ntiderivtive of sin x is cos x so π 0 sin x dx = cos π + cos 0 = ( 1) + 1 = 2
Why is the FTC true? d x f (t) dt = f (x) dx We wnt to show tht d dx F (x) = f (x).
Why is the FTC true? d x f (t) dt = f (x) dx We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h
Why is the FTC true? d x f (t) dt = f (x) dx We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h 1 = lim h 0 h [ x+h x ] f (t) dt f (t) dt
Why is the FTC true? d x f (t) dt = f (x) dx We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h 1 = lim h 0 h 1 = lim h 0 h [ x+h x+h x x ] f (t) dt f (t) dt f (t) dt
Why is the FTC true? When h is very smll x+h x f (t)
Why is the FTC true? When h is very smll x+h x f (t) hf (x)
Why is the FTC true? We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h 1 = lim h 0 h 1 = lim h 0 h [ x+h x+h x x ] f (t) dt f (t) dt f (t) dt
Why is the FTC true? We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h 1 = lim h 0 h 1 = lim h 0 h = lim h 0 1 [ x+h x+h x hf (x) h x ] f (t) dt f (t) dt f (t) dt
Why is the FTC true? We wnt to show tht d dx F (x) = f (x). d F (x + h) F (x) F (x) = lim dx h 0 h 1 = lim h 0 h 1 = lim h 0 h h 0 1 [ x+h x+h x = lim hf (x) h = f (x) x ] f (t) dt f (t) dt f (t) dt
A consequence (corrollry) Corollry For ny ntiderivtive F (x) of f (x) f (x) dx = F (b) F ()
A consequence (corrollry) Corollry For ny ntiderivtive F (x) of f (x) f (x) dx = F (b) F () Why? Well F (x) = x f (t) dt + C for some nd C. So F (b) F () = = f (t) dt + C f (t) dt f (t) dt C
The indefinite integrl We lso use the following nottion for the generl ntiderivtive: Definition f (x) dx = F (x) + C
The indefinite integrl We lso use the following nottion for the generl ntiderivtive: Definition f (x) dx = F (x) + C Exmple sin(x) x dx = cos(x) 1 2 x 2 + C
Integrtion by substitution As we hve seen, some ntiderivtives re difficult to guess,
Integrtion by substitution As we hve seen, some ntiderivtives re difficult to guess, especilly if it involves reversing the chin rule.
Integrtion by substitution As we hve seen, some ntiderivtives re difficult to guess, especilly if it involves reversing the chin rule. We solve this by introducing new vrible.
Integrtion by substitution As we hve seen, some ntiderivtives re difficult to guess, especilly if it involves reversing the chin rule. We solve this by introducing new vrible. Substitution Suppose u = g(x), then f (g(x))g (x) dx = f (u) du
Integrtion by substitution As we hve seen, some ntiderivtives re difficult to guess, especilly if it involves reversing the chin rule. We solve this by introducing new vrible. Substitution Suppose u = g(x), then f (g(x)) du dx dx = f (g(x))g (x) dx = f (u) du
Exmple 1 Question 4x x 2 + 1 dx
Exmple 1 Question 4x x 2 + 1 dx Solution We use the substitution u = x 2 + 1, so du dx = 2x, we cn write the integrl 2 x u 2 + 1 2x dx = 2 du
Exmple 1 Question 4x x 2 + 1 dx Solution We use the substitution u = x 2 + 1, so du dx = 2x, we cn write the integrl 2 x u 2 + 1 2x dx = 2 du = 2 2 3 u 3 2 + C
Exmple 1 Question 4x x 2 + 1 dx Solution We use the substitution u = x 2 + 1, so du dx = 2x, we cn write the integrl 2 x u 2 + 1 2x dx = 2 du = 2 2 3 u 3 2 + C = 4 3 (x 2 + 1) 3 2 + C
Integrtion by substitution (definite integrls) Substitution for definite integrls Suppose u = g(x), then f (g(x))g (x) dx = g(b) g() f (u) du
Integrtion by substitution (definite integrls) Substitution for definite integrls Suppose u = g(x), then f (g(x))g (x) dx = g(b) g() f (u) du Exmple 1 0 4x x 2 + 1 dx = 2 2 1 u du
Integrtion by substitution (definite integrls) Substitution for definite integrls Suppose u = g(x), then f (g(x))g (x) dx = g(b) g() f (u) du Exmple 1 0 4x x 2 + 1 dx = 2 = 2 2 1 [ 2 3 u 3 2 u du ] 2 1
Integrtion by substitution (definite integrls) Substitution for definite integrls Suppose u = g(x), then f (g(x))g (x) dx = g(b) g() f (u) du Exmple 1 0 4x x 2 + 1 dx = 2 2 1 u du [ ] 2 2 = 2 3 u 3 2 1 ( ) 2 = 2 3 2 3 2 2 3 1 3 2 = 4 3 (2 2 1)