AEA integration page!
6. (a) Use the substitution u= √t to show that
Int
d=4-4√x +2√x lnx x>1
t
(5)
(b) The function g is such that
S g(t) dt =x-√x Inx-1 x >1
(i) Use differentiation to find the function g.
م61
(i) Evaluate () dr and simplify your answer.
(5)
(c) Find the value of x (where x > 1) that gives the maximum value of
x+1 Int
dt
J. 2
(7)
6. (a) Show that
1
du In(u+√u²-1)= √u²-1
(2)
1
(b) Use the result from part (a) and the substitution x +3 =
t
to find
1
=dx
J(x+3)√2x+7
(6)
1
(c) Express in partial fractions.
2x² +13x +21
(2)
(d) Find
1
dx
(2x² + 13x+ 21)√2x +7
giving your answer in the form Inr-s where r and s are rational numbers.
(6)
, 1J
S
(60 ==
压 dt= L) n da du.
=S 2)
dt = 2Jt du = 2u du
= 2√ال In(w) du u In/u)u
2
2 du
=2 u lnlu) -24
= 2 [x (n(x) 4 [x
- - 2ln() +4
= 4-4Jx + 2kln(x)
(6A let Glt) =jglt)dt Sf glt) dt = [Gto];= 66)- 6()
G(x) = fg lx) dx
(f gat) = (66)-60) G(1) = 1 - Tiln1-1 = 0
= 9(x)
1-
S geabdet S sthldt - 4+104+1
bil
)
=
16 - Diln6-1 -
=
12-4lnl6 + 21n4
= 12-4 In 2t + 21n 2
= 12-16 [n2+ 4 In2
=12-12 1n2
, xtl
let alz) = Jx
dt
-
Int
Q'(x) = [n(xti) (n(e)
2(m) - 20
we want max a(x), 3o find a(x) =0
In(x+) In(x)
2x
7 ((nlkr)- 2ine) = 0
≠0 for allx>
{n(xt1) -210(k] = 0
(n(x+1) = (n(x)
x²-x-1 = 0
土J-4(1)(-)
x=
2
1
2
+5
SO
only
6. (a) Use the substitution u= √t to show that
Int
d=4-4√x +2√x lnx x>1
t
(5)
(b) The function g is such that
S g(t) dt =x-√x Inx-1 x >1
(i) Use differentiation to find the function g.
م61
(i) Evaluate () dr and simplify your answer.
(5)
(c) Find the value of x (where x > 1) that gives the maximum value of
x+1 Int
dt
J. 2
(7)
6. (a) Show that
1
du In(u+√u²-1)= √u²-1
(2)
1
(b) Use the result from part (a) and the substitution x +3 =
t
to find
1
=dx
J(x+3)√2x+7
(6)
1
(c) Express in partial fractions.
2x² +13x +21
(2)
(d) Find
1
dx
(2x² + 13x+ 21)√2x +7
giving your answer in the form Inr-s where r and s are rational numbers.
(6)
, 1J
S
(60 ==
压 dt= L) n da du.
=S 2)
dt = 2Jt du = 2u du
= 2√ال In(w) du u In/u)u
2
2 du
=2 u lnlu) -24
= 2 [x (n(x) 4 [x
- - 2ln() +4
= 4-4Jx + 2kln(x)
(6A let Glt) =jglt)dt Sf glt) dt = [Gto];= 66)- 6()
G(x) = fg lx) dx
(f gat) = (66)-60) G(1) = 1 - Tiln1-1 = 0
= 9(x)
1-
S geabdet S sthldt - 4+104+1
bil
)
=
16 - Diln6-1 -
=
12-4lnl6 + 21n4
= 12-4 In 2t + 21n 2
= 12-16 [n2+ 4 In2
=12-12 1n2
, xtl
let alz) = Jx
dt
-
Int
Q'(x) = [n(xti) (n(e)
2(m) - 20
we want max a(x), 3o find a(x) =0
In(x+) In(x)
2x
7 ((nlkr)- 2ine) = 0
≠0 for allx>
{n(xt1) -210(k] = 0
(n(x+1) = (n(x)
x²-x-1 = 0
土J-4(1)(-)
x=
2
1
2
+5
SO
only
