Mathias André
Maxime Tô
M1 EPP 2010
Maths Summer Camp
Exercises : Sessions 7 and 8
Master Economics and Public Policies, Maths Summer Camp
"Usual" Primitives
Calculate (and learn) the following primitives :
R dx
1.
x2 −a2
ax
2.
R
e dx
3.
R
ax dx with a > 0
4.
R
dx
x−a
5.
R
ln xdx
6.
R
(x − a)α dx with α 6= −1
7.
R
√ dx√
x− a
with a > 0
Primitives
Calculate the following primitives :
R
1. x ln(x2 + 1)dx
R
2. x ln xdx
R 2x 2
3. e (x − x + 3) dx
R dx
4. 3x+1
R
2
)
5. 2+ln(x
x
R 3√
6. x 1 + x2 dx
Exercises
Rb
(x − a)3 (b − x)4 dx
R a x ln x
2. Calculate I = 1 (1+x2 )2 dx (use a change in variable)
1. Express I =
a
a
3. f ∈ C 2 ([a, b], R). Show that :
Z
b
f (x)dx =
a
1
b−a
[f (a) + f (b)] +
2
2
Pn−1
Z
b
(x − a)(x − b)f 00 (x)dx
a
1
4. α > 0, β > 0. Consider un = k=0 αn+βk
. Find lim un when n → +∞.
R 1 xn −x2n
5. In = 0 1−x dx. Check that In is well defined and find lim In when n → +∞.
1
Solutions : Sessions 5 and 6
Master Economics and Public Policies, Maths Summer Camp
"Usual" Primitives
All the primitives are up to an additive constant :
R
1
x−a
1. x2dx
−a2 = 2a ln | x+a |
R ax
ax
2. e dx = ea
R
x
3. ax dx = lna a with a > 0
R dx
4. x−a
= ln |x − a|
R
5. ln xdx = x ln x − x
R
α+1
6. (x − a)α dx = (x−a)
with α 6= −1
α+1
R dx
√
√
√
√
7. √x−√a = 2( a ln( x − a + x) with a > 0
Primitives
All the primitives are up to an additive constant :
R
1. x ln(x2 + 1)dx = 21 ((x2 + 1) ln(x2 + 1) − x2 ). (by part or change of variable)
R
2
2. x ln xdx = x4 (2 ln x − 1). (by part)
R
3. e2x (x2 − x + 3) dx = 12 e2x (x2 − 2x + 4). (by part)
R dx
4. 3x+1
= 13 ln(3x + 1)
R
R
2
)
= 14 ln2 (x2 ) + 2 ln x. ( u0 (x)u(x)dx = u2 (x))
5. 2+ln(x
x
R √
3
1
6. x3 1 + x2 dx = 15
(1 + x2 ) 2 (3x2 − 2). (t = 1 + x2 )
Exercises
1. I =
(b−a)8
280
(change of variable y = b − x then z =
2. I = 0 with t =
1
x
y
b−a ).
with it the interval of integration remain unchanged).
3. Integrate by parts the right handside.
Pn−1 1
α
4. It is a Rieman’s sum un = n1 k=0 α+β
k . Then un → ln(1 + β ) when n → +∞.
n
P2n−1 R
P2n−1
n
−x2n
5. x 1−x
= k=n xk implies that In is well defined (only one problem around 1) and In k=n xk dx =
P2n−1 k Pn−1 1
k=n x =
k=0 n+k . Using previous exercise, In → ln 2 when n → +∞.
2