Scarica Esercizi di Calcolo Integrale - Universit√á Bocconi e più Esercizi in PDF di Matematica Generale solo su Docsity!
Exercises 1. Integral calculus
Bocconi University
a.a. 2018/
R 1
1
10 x 2
dx =
a 0 b
c
d none of the proceeding
R (^) e
2
1 x ln x dx =
a 2 ln 2 3 2 b 4 ln 2 c 4 ln 2 + 3 2 d none of the proceeding
- The antiderivative of f (x) = 1 (1+x)^3
passing through the point (0; 1) is:
a
1 2(1+x)
2 +^
1 2 b^ ^
1 (1+x) 4 + 2^ c^ ^
1 2(1+x)
2 +^
3 2 d^ none of the proceeding
- Solve
Z^2
1
x+ x^2 +2x dx :
a 1 2 ln 1 3 b 3 2 ln 2 1 2 ln 3 c 0 d none of the proceeding
- Compute the following indeÖnite integrals:
a)
R
x
p x 1 dx; b)
R
(x + 3) sin xdx;
c)
R
ln 2 xdx; d)
R
1 1+ex^ dx.
e)
R
2 x(1+ln x)^2 dx ; f)
R p x+ x dx.
R (^) ln(1+px) p x dx =
a) 2 x log(1 +
p x) 2
p x + 2 log(1 +
p x) + k
b) 2
p x log(1 +
p x) 2
p x + 2 log(1 +
p x) + k
c)
p x log(1 +
p x) + k
d) none of the proceeding
- Find an antiderivative G(x) of the following functions that satisfy the condi-
tion:
i) f (x) = x e x ; 2G(0) = G(1).
ii) f (x) = x 1 x+
; 2G(1) + G(2) = 4.
R
4 2 x 2 e x dx =
a) 5 e 2 2 e 2 b) 0 c) 10 e 4 2 e 2 d) 7 e 2
- Compute
R 2
0 sin
p xdx:
- Find the values of a 2 R such that
R
a 0 (x 3 x 2 )dx = 0:
- Compute: Z (^2)
1
f (x)dx; con
f (x) =
x x 1 ; x < 0 ex ex+ ; x 0
- Compute the deÖnite integrals on the indicated interval.
(a) f (x) = sin x sin(cos x);
4
(b) f (x) =
5 p x + 2; x 0
ln(x + 1); x > 0
; [ 3 ; 1].
R
dx x(x^2 x+12)
a) diverges b) converges c) 9 / d) none of the proceeding
R + 1
0
sin x+ ex+e x^ dx:
a) converges b) diverges c) 9 / d) none of the proceeding
R + 1
1
p^ x+ln^ x x+x^2 1 dx:
a) converges b) diverges c) 9 / d) none of the proceeding
- Find the values of a 2 R such that the following generalized integral con-
verges/diverges: Z (^) + 1
2
x ln
a x
dx
- Compute
+R 1
10
xe 1 2 x
2 dx.
- Let f (x) =
x; x < 1
x x 2 ; x 1
write the expression of F (x) =
R
x 0 f (t)dt.
ii) G(x) =
R
x 1 x+ dx =
R
x+2 2 1 x+ dx =
R
x+ x+ dx +
R
3 x+ dx =
R
1 dx 3
R
1 x+ dx = x 3 ln jx + 2j + k
G(1) = 1 3 ln 3 + k
G(2) = 2 3 ln 4 + k
2 G(1) + G(2) = 4 : 2 6 ln 3 + 2k + 2 3 ln 4 + k = 4
k = 2 ln 3 + ln 4 = ln 9 + ln 4 = ln 36
F (x) = x 3 ln jx + 2j + ln 36
- c)
R
x 2 e x dx = e x x 2
R
2 xe x dx = e x x 2 2 e x x+
R
2 e x dx = e x x 2 2 e x x + 2e x
R (^4)
2 x 2 e x dx =
e x x 2 2 e x x + 2e x
2 = 16e 4 8 e 4
- 2e 4 (4e 2 4 e 2
- 2e 2 ) =10e 4 2 e 2
- By substitution
p x = t; x = t 2 ; dx = 2tdt
R
2 t sin tdt = 2 t cos t+
R
2 cos tdt =
2 t cos t + 2 sin t R (^) 2
0 sin
p xdx = [ 2
p x cos
p x + 2 sin
p x]
^2 0 = 2
R (^) a
0 (x 3 x 2 )dx =
h x^2 2 x 3
ia
0
a^2 2 a 3
a^2 2 a 3 = 0, then a = 0 or a = 1 2
R
2 1 f (x)dx =
R
0 1
x x 1 dx+
R
2 0
ex ex+ dx = [x + ln jx 1 j]
0 1 + [ln (e
x
2 0 =
1 + ln(e 2
- The integrals are:
(a)
R
4 0 sin^ x^ ^ sin (cos^ x)^ dx= [cos (cos^ x)]^
4 0 = cos
(^) p 2 2
cos (1)
(b)
R 0
3
5 p x + 2dx+
R 1
0 ln(x^ + 1)dx=^
5 3
5 p 2 + 2 ln 2
11 6
- b) for x! + 1 1 x(x^2 x+12)
1 x^3
- a) for x! + 1 0
sin x+ ex+e x^
3 ex+e x^
3 ex^ ^
3 x^10
- b) for x! + 1 px+log^ x x+x^2 1
x x^2
1 x
- If a = 1 we have:
R
1 x ln x dx = lim k!+ 1
R^ k
2
1 x ln x dx = lim k!+ 1
[ln jln xj]
k 2 =
= lim k!+ 1
[ln jln kj ln jln 2j] = + 1
If a 6 = 1 we have:
R + 1
2
1 x lna^ x dx =
R + 1
2
ln a^ x x dx = lim k!+ 1
R^ k
2
ln a^ x x dx = lim k!+ 1
h ln a+1^ x a+
ik
2
= lim k!+ 1
h ln a+1^ k a+
ln a+1^2 a+
i
lim k!+ 1
h 1 ( a+1) lna ^1 k
1 ( a+1) lna ^1
i
if a 1 > 0 (a > 1) the integral converges because 1 ( a+1) lna ^1 k
if a 1 < 0 (a < 1) the integral diverges because 1 ( a+1) lna ^1 k
ln^1 a^ k ( a+1)
We can state that the integral converges if a > 1 and it diverges if a 1.
- We get:
+R 1
10
xe 1 2 x
2 dx = lim k!+ 1
R^ k
10
xe 1 2 x
2 dx = lim k!+ 1
h e 1 2 x
2
ik
10
= lim k!+ 1
h e 12 k^2
i = e 50 (It converges)
- If x < 1 ; F (x) =
R (^) x 0 tdt^ =^
x^2 2
If x 1 ; F (x) =
R 1
0 tdt +
R (^) x
1 t t 2 dt = 1 2
x^2 2
x^3 3
1 2
1 3
F (x) =
x^2 2 x < 1 x^2 2 ^
x^3 3 +^
1 3 x^ ^1
- If x < 0 ; F (x) =
R (^) x 2 f (t)dt =
R 2
x f (t)dt =
R 0
x
1+2t 1+t+t^2 dt +
R 2
0 e t dt
ln 1 + t + t 2
x
e t
0
ln 1 ln 1 + x + x 2
= ln 1 + x + x 2 e 2
If x 0 ; F (x) =
R (^) x 2 e t dt =
e t x
2
= e x e 2
F (x) =
ln 1 + x + x 2 e 2
e x e 2 x 0
- The function is negative for 0 < x < 1 and it is positive for 1 < x < 2 , so we
break the deÖnite integral:
Area =
R 1
0
x 2 x
dx +
R 2
1
x 2 x
dx =
h x^3 3
x^2 2
i 1
0
h x^3 3
x^2 2
i 2
1
1 3
1 2
8 3
4 2
1 3
1 2
1 6
8 3
4 2
1 3
1 2
1 6
8 3
4 2
1 3
1 2
