HOAT DONG 1
• Thuc hien .Q, 1 trong 5'
GV treo hoac chieu hinh 45, hinh 46 trong SGK.
cau 1)
Hoat dgng cua GV Cau hdi 1
'Tfnh chieu cao ciia hinh thang.
Cau hdi 2
Tfnh chieu dai hai day cua hinh thang.
Cau hdi 3
Tfnh dien tfch hinh thang.
Hoat dgng ciia H S Ggi y tra loi cau hdi 1
h = 5 - l = 4 .
Ggi y tra loi cau hdi 2
Chieu dai hai day la f(l) -
v a f ( 5 ) = l l .
Goi y tra Idi cau hdi 3
-- 3
cau 2)
Hoat dgng cua G V Cau hdi 1
Tinh chieu cao cua hinh thang.
Cau hdi 2
Tfnh chidu dai hai day ciia hinh thang.
Cau hdi 3
Tfnh dien tfch hinh thang.
Hoat dgng ciia HS Ggi y tra Idi cau hdi 1
h = 5 - l = 4 .
Ggi y tra Idi cau hdi 2
Chieu dai hai day la f(l) - 3
va f(t) = 2t + 1
Ggi y tra loi cau hdi 3
Sit)=^^^y\t l) = t' + t 2, (adsbygoogle = window.adsbygoogle || []).push({});
ts[l- 5].
cau 3)
Hoat dgng cua G V Cau hdi 1
Chiing minh S(t) la nguyen ham cua f(t) = 2x + 1.
Cau hdi 2
Hoat dgng ciia HS Ggi y tra loi cau hdi 1
Vi S'it) = 2t+l,t e[l ; 5], nen Sit) la mdt nguyen ham ciia fit) = 2t + I
Chirng minh S = S(5) - S(l). cJ = S ( 5 ) - 5 ( l ) = 2 8 - 0 = 28. • Tiep theo GV sii dung hinh 47 dé mo ta dien tfch hinh thang cong. • GV neu dinh nghia hinh thang cong.
Cho hdm sd y = fix) lien tuc, khong doi dau tren doan /a; b/. Mot hinh phang gidi hgn bdi đ thi cua hdm sd fix), true hodnh vd hai dudng thdng x = a, x = b duac ggi Id hinh thang cong.
HỊ Hay neu mdt vai vf du ve hinh thang cong. • Thuc hien vf du 1 trong 5'
Hoat dgng ciia GV Cau hdi 1 Tfnh dien tfch hinh MNPQ. Cau hdi 2 Tfnh dien tfch hinh M N E F Cau hdi 3
Tfnh dien tfch hinh thang cong M N Q E .
Cau hdi 4
Chiing minh S'(x) = x^
Hoat dgng ciia H S Ggi y tra Idi cau hdi 1
SMNPQ = h. f(x) = hx^
Ggi y tra Idi cau hdi 2
SMNPQ = h. f(x+h) = h(x + h)
Ggi y tra loi cau hdi 3
S(x + h) - S(x).
Ggi y tra Idi cau hdi 4
GV hudng đn HS chiing minh tuong tu SGK
• GV neu dien tfch hinh thang cong bdt k i :
Vdi moi x e [a;b], ki hieu S(x) la dien tich cua phdn hinh thang cong do nam giiia hai dudng thdng vuong goc vdi Ox Idn litdt tgi a vd X. Ta cdng chiing minh duac S(x) la nguyen hdm cda f(x) tren dogn [a ; bJ. Gia sicF(x) cdng Id mot nguyen hdm cua f(x) thi co mot hang sdC sao cho Six) = Fix) + C.
Vi S(a) = 0 nen F(a) + C = 0 hay C = -F(a).Vdy Six) = Fix) - Fia).
Thay x = b vao dang thitc tren, ta co dien tich cua hinh thang cdn tim la Sib) = Fib)-Fia).
HOAT DONG 2
2. Dinh nghia tich phan
• Thuc hien "^t 2 trong 5'
Hoat dgng ciia GV Cau hdi 1 (adsbygoogle = window.adsbygoogle || []).push({});
Gia su F(x) va G(x) la hai nguydn ham ciia ham f(x). Chumg minh F(x) - G(x) = C.
Cau hdi 2
Chumg minh F(a) - F(b) = G(a) - G(b).
Hoat dgng ciia HS Ggi y tra loi cau hdi 1
HS tu chumg minh theo dinh nghia nguydn ham.
Ggi y tra loi cau hdi 2
HS tu chumg minh
• GV neu dinh nghia
Cho fix) la ham sd lien tuc tren doan [a ; 6]. Gia sir Fix) la mdt nguyen
ham ciia/"(x) tren doan [a ; 6].
Hieu sd Fib) - Fia) dugc ggi la tich phan tii a den b (hay tich phan xac
6
dinh tren doan [a ; b]) ciia ham so fix), ki hieu la [/'(x)dx.
lfix)dx = Fix)\l=Fib)-Fia).
Ta ggi j la dau tich phdn vdi a la can dudi va b la can tren, fix)dx la
a
a H2. Chiing minh Jf (x)dx = 0. a b a H3. Chiing minh ff (x)dx = - jf (x). a b
• GV neu chii y trong SGK
• Thuc hien vf du 2 trong 5'. GV cd the láy vf du khac
cau ạ
Hoat dgng cua GV Cau hdi 1
Tim nguyen ham F(x) ciia ham sd y = 2x. Cau hdi 2 2 Tfnh J2xdx. 1 Hoat dgng cua HS Ggi y tra Idi cau hdi 1
F(x)= f2xdx = x^
Ggi y tra Idi cau hdi 2
HS tu tfnh.
caub.
Hoat dgng cua GV Cau hoi 1
Tim nguydn ham F(t) cua ham sd'y = - . Cau hdi 2 2 Tfnh f - d t . 1 Hoat dgng cua HS Ggi y tra loi cau hdi 1 F(t)= [-dt = lnt.
•"t
Ggi y tra Idi cau hoi 2 HS tu tinh. (adsbygoogle = window.adsbygoogle || []).push({});
• GV neu chu y :
b) Y nghia hinh hgc cua tich phdn : Neu hdm sdf(x) lien tuc vd khong b
dm tren dogn (a ; b], thi tich phdn fix) dx Id dien tich S cua hinh a
thang cong gidi hgn bdi đ thi cda f(x), true Ox vd hai dudng thdng X = a,x = b. Vay
H4. Tfnh cac dien tich hinh thang d hoat dgng 1 bang tich phan.
HS.Tinh jx^dx.
1
2
H6. Tinh [inxdx.
HOAT DONG 3