Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28
Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)] x=b x= = F(b) F() where F is ny ntiderivtive of f on (, b).
Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)] x=b x= = F(b) F() where F is ny ntiderivtive of f on (, b). Question: Cn we evlute the definite integrl x 2 dx?
Answer We cnnot use the Fundmentl Theorem of Clculus to evlute x 2 dx since the integrnd hs discontinuity t x =. If we try to evlute it using the Fundmentl Theorem of Clculus we get [ x 2 dx = ] x= = 2 x x= result which is impossible since /x 2 > for x < nd < x.
Improper Integrls Extr cre must be exercised when ttempting to evlute definite integrls for which the intervl over which we integrte is of infinite length (Type ),
Improper Integrls Extr cre must be exercised when ttempting to evlute definite integrls for which the intervl over which we integrte is of infinite length (Type ), nd/or the integrnd possesses isolted discontinuities within the integrtion intervl (Type 2).
First Type (Type ) Definition If f is continuous on [, ) we define the improper integrl R f (x) dx f (x) dx. R If f is continuous on (, ] we define the improper integrl f (x) dx = lim R R f (x) dx. If the limit is L (finite) we sy the improper integrl converges, otherwise we sy it diverges.
Exmples Determine if the following improper integrls converge or diverge.. 2. 3. 5 5 e x dx x dx x 2 dx
e x dx e x dx R This improper integrl converges. R e x dx [ e x ] x=r R x= [ e R + e ] R = e
5 x dx 5 dx x = R lim R 5 x dx [ln x ]x=r x=5 R [ln R ln 5] = This improper integrl diverges. R
5 x 2 dx 5 dx x 2 R R R R 5 [ x x 2 dx ] x=r x=5 [ R + 5 ] = 5 This improper integrl converges.
Intervl (, ) Definition If f is continuous on (, ) then f (x) dx = for ny constnt. We sy diverges. f (x) dx nd f (x) dx + f (x) dx f (x) dx converges if both f (x) dx converge, otherwise f (x) dx
Exmple Determine if the following improper integrl converges. e x dx + e x
Solution ( of 2) = e x + e x dx e x + e R R R R dx + x e x + e e x + e x dx + lim e x e 2x dx + lim + S S S dx x S e x dx + e x e x e 2x + dx Use integrtion by substitution with u = e x nd du = e x dx.
Solution (2 of 2) e x + e R R = R dx x e R [ π 4 ] + e S u 2 du + lim + S u 2 + du [ ] u= [ ] u=e tn u + lim tn S u u=e R S u= [ tn tn e R] + lim S [ π 2 π ] = π 4 2 This improper integrl converges. [ tn e S tn ]
Grphicl Approch Suppose f nd g re two continuous functions defined on [, ) nd such tht f (x) g(x) for ll x. y f(x) g(x) x
Grphicl Approch Suppose f nd g re two continuous functions defined on [, ) nd such tht f (x) g(x) for ll x. y f(x) g(x) x If f (x) dx diverges, wht bout g(x) dx?
Grphicl Approch Suppose f nd g re two continuous functions defined on [, ) nd such tht f (x) g(x) for ll x. y f(x) g(x) x If If f (x) dx diverges, wht bout g(x) dx converges, wht bout g(x) dx? f (x) dx?
Comprison Test Theorem (Comprison Test) Suppose tht f nd g re continuous on [, ) nd f (x) g(x) for ll x.. If 2. If g(x) dx converges, then f (x) dx diverges, then f (x) dx converges. g(x) dx diverges.
Exmples Determine if the following improper integrls converge or diverge.. 2. 3. x 2 2 x 4 + 3 dx + sec 2 x x e x+ dx dx
x 2 2 x 4 + 3 dx Note tht so x 2 2 x 4 + 3 < x 2 x 4 + 3 < x 2 x 4 = x 2 x 2 2 x 4 + 3 dx < x 2 dx R R x 2 dx [ ] x=r R x x= [ R ] + =. R The originl improper integrl converges.
+ sec 2 x x dx Note tht so + sec 2 x x > x + sec 2 x x dx > R x dx R x dx [ln x ]x=r x= R [ln R ln ] =. R The originl improper integrl diverges.
e x+ dx Note tht so e x+ dx > e x+ = e e x > e x R e x dx R e x dx R [ex ] x=r x= [ ] e R =. R The originl improper integrl diverges.
Second Type (Type 2) Definition If f is continuous on the intervl [, b) nd f (x) s x b, the improper integrl of f on [, b] is b f (x) dx = R lim f (x) dx. R b If f is continuous on the intervl (, b] nd f (x) s x +, the improper integrl of f on [, b] is b f (x) dx = b lim f (x) dx. R + R If the limit is L (finite), we sy the improper integrl converges, otherwise we sy it diverges.
Exmples Determine if the following improper integrls converge or diverge.. 2. 3. 4 π/2 2x x 2 dx 3 x dx tn x dx
4 2x x 2 dx 4 2x 4 2x x 2 dx R + R x 2 dx [ ] x=4 ln x 2 R + x=r ] ln 5 ln R 2 = R + [ This improper integrl diverges.
3 x dx 3 dx x /3 dx x R + R [ ] 3 x= R + 2 x 2/3 x=r [ 3 R + 2 3 ] 2 R2/3 = 3 2 This improper integrl converges.
π/2 tn x dx π/2 tn x dx R R π/2 R R π/2 This improper integrl diverges. tn x dx sin x cos x dx [ ln cos x ]x=r R π/2 x= [ ln cos R + ln cos ] R π/2 ( ) = lim ln cos R R π/2 =
Discontinuity in (, b) Definition Suppose f is continuous on [, b] except t some c (, b) nd f (x) s x c. The improper integrl is If c b b f (x) dx = f (x) dx = L nd b c c b f (x) dx + f (x) dx. c f (x) dx = L 2 the improper integrl f (x) dx converges to L + L 2. If either of the improper integrls c diverges s well. f (x) dx or b c f (x) dx diverges then b f (x) dx
Exmples Determine if the following improper integrls converge or diverge.. 2. 4 3 x dx 2x x 2 dx
3 x dx 3 x dx = R R R 3 dx + x R 3 x dx x /3 dx + lim [ ] 3 x=r 2 x 2/3 [ 3 2 R2/3 3 2 x= S + + lim S + ] + lim S + S x /3 dx [ ] 3 x= 2 x 2/3 x=s [ 3 2 3 2 S2/3 ] = 3 2 + 3 2 = This improper integrl converges.
4 2x x 2 dx 4 2x x 2 dx = 2x 4 x 2 dx + 2x x 2 dx We hve lredy shown tht the second integrl on the right-hnd side of the eqution diverges, thus this improper integrl diverges.
Homework Red Section 7.8 Exercises: WebAssign/D2L