Review: The Riemann Integral Review: The definition of R b

Transcription

1 eview: The iemnn Integrl eview: The definition of b f (x)dx. For ontinuous funtion f on the intervl [, b], Z b f (x) dx lim mx x i!0 nx i1 f (x i ) x i. This limit omputes the net (signed) re under the grph of f (to the x-xis) on the intervl [, b]. Figure: The Limit Proess behind the iemnn Integrl We hop the intervl [, b] inton subintervls [x 0, x 1 ], [x 1, x ],...,[x n 1, x n ], where x 0 nd x n b. The width of the i-th subintervl [x i 1, x i ]is x i x i x i 1.Thisisthebse width of retngle built upon the i-th subintervl. Wht bout its height? For i 1,,...,n, x i is smple point tken from the i-th subintervl. The funtion vlue f (x i )istheheight of the retngle built upon the i-th subintervl. The re of the i-th retngle is f (x i ) x i. Mth 67 (University of Clgry) Fll 015, Winter / 9

2 eview: The iemnn integrl Figure: The Limit Proess behind the iemnn Integrl We now tke the sum of the res of ll n retngles. The nswer pproximtes the re under the grph of f (nd bove the x-xis) on the intervl [, b]. We now tke lots of retngles, ll being thin. This is expressed in the limit opertion on the right-hnd side of Z b nx f (x) dx f (x i ) x i. lim mx x i!0 iemnn sums (nd integrls) rebouthoppingthingsintolittlepiees. i1 Mth 67 (University of Clgry) Fll 015, Winter 016 / 9

3 Double integrls Suppose z f (x, y) isontinuousontheretngulrregion :[, b] [, d], where {(x, y) : pple x pple b, pple y pple d}. Figure: Wht is the volume under the grph of z f (x, y)? We wnt to find the net (signed) volume V under the grph of f (to the xy-plne) on the region. Mth 67 (University of Clgry) Fll 015, Winter / 9

4 Double integrls Figure: Volume of smple retngulr tower f (x ij ) x i y j Chop into smll retngulr piees using grid defined by prtition of [, b] ndprtitionof[, d]: [, b] :[x 0, x 1 ], [x 1, x ],...,[x m 1, x m ] [, d] :[y 0, y 1 ], [y 1, y ],...,[y n 1, y n ], where x 0, x m b, y 0, y n d. The retngulr piees re doubly indexed by i nd j, withwidthndlengthgivenby x i x i x i 1 nd y j y j y j 1,wherei {1,,...,m} nd j {1,,...,n}. Itsreis A ij x i y j. In the retngulr piee indexed by (i, j), tke smple point (x ij ), nd build retngulr tower (or box) with height equl to f (x ij ). The volume of the tower is equl to V ij f (x ij ) A ij. Mth 67 (University of Clgry) Fll 015, Winter / 9

5 Double integrls (CONTINUED) Add up the volumes of ll mn towers. The vlue of this double sum pproximtes the volume under the grph of f (nd bove the xy-plne) on the region. V mx i1 nx j1 f (x ij ) A ij. We now bring the number of piees to infinity, with eh x i nd y j pprohing zero. This is formlly limitproess. The net volume under the grph of f (nd bove the xy-plne) on is equl to the following double integrl defined by mx nx f (x, y) da f (x ij ) A ij. lim mx x i, y j!0 i1 j1 Mth 67 (University of Clgry) Fll 015, Winter / 9

6 Iterted integrls nd Fubini s Theorem How to ompute the vlue of double integrl? One n use the following Fubini s Theorem Let f be ontinuous on the retngulr region [, b] [, d]. Then Z d Z b Z b Z d f (x, y) da f (x, y) dx dy f (x, y) dy dx. The integrls in the middle nd on the right re lled iterted integrls. To ompute the iterted integrl d b f (x, y) dx dy, I strt on the inside nd evlute b f (x, y) dx, holdingy onstnt. I then integrte d... dy (the outside integrl) s usul. Figure: Two orders of integrtion in Iterted integrls Mth 67 (University of Clgry) Fll 015, Winter / 9

7 Evluting double integrls by iterted iterted integrls To help ourselves keep trk of whih numbers to substitute into whih vrible, we sometimes write Z xd Z xb f (x, y) dx dy for d Exmple Compute b f (x, y) dx dy. y x (6 y x) da where is the retngulr region defined by 0 pple x pple 1nd0pple y pple. Solution: (6 y x) da Z 0 Z y y0 Z 1 0 Z y (6 y x) dx dy y0 6x yx x x1 dy x0 6 (1) y (1) (1) 6 (0) y (0) (0) dy Z y y0 5y (6 y 1) dy! y y y0 10 Z y y0 (5 y) dy!! 0 8. Mth 67 (University of Clgry) Fll 015, Winter / 9

8 Using the order dydx (6 y x) da Z 1 0 Z 0 (6 y x) dy dx Z x1 x0 Z x1 x0 6y 6 () y! y xy dx y0! x () 6 (0) 0 x (0)!! dx Z x1 x0 (1 4x) dx 10x x x1 x0 10 (1) Z x1 x0 (10 4x) dx To find the sum of n rry of numbers, we n dd rows then olumns or olumns then rows: 1 0! ! 4 7 3! 14 # # # # 8 6 5! 19 Mth 67 (University of Clgry) Fll 015, Winter / 9

9 Further exmples/exerises 1 Find the following double integrls. Try both orders of iterted integrtion. I (x +3y) da, whered [1, 3] [, 4]. D h I (e x + x os y) da, whered [ 1, 1] 0, i. D x 3 + y 3 I D xy da, whered is the retngle defined by 1 pple x pple, 1 pple y pple 3. Show tht if f is ontinuous on [, b] ndg is ontinuous on [, d], then Z b Z d f (x)g(y) da f (x) dx g(y) dy. [,b] [,d] Mth 67 (University of Clgry) Fll 015, Winter / 9