> 1

\

/---------------------------------

x

|

exp(x)

|

| normcdf(x) |

|-------

|

------------

|

------------|

| -4.000|

0.0183|

0.0000|

| -3.500|

0.0302|

0.0002|

| -3.000|

0.0498|

0.0013|

| -2.500|

0.0821|

0.0062|

| -2.000|

0.1353|

0.0228|

| -1.500|

0.2231|

0.0668|

| -1.000|

0.3679|

0.1587|

| -0.500|

0.6065|

0.3085|

|

0.000|

1.0000|

0.5000|

|

0.500|

1.6487|

0.6915|

|

1.000|

2.7183|

0.8413|

|

1.500|

4.4817|

0.9332|

|

2.000|

7.3891|

0.9772|

|

2.500|

12.1825|

0.9938|

|

3.000|

20.0855|

0.9987|

|

3.500|

33.1155|

0.9998|

|

4.000|

54.5982|

1.0000|

\---------------------------------

/

номер вариант

функция 1

функция 2

1

x

−

y2

y=−∞∫

2

f (x )=

e

dy

1

f (x )=sin( x)

√

(2 π)

(normcdf в MATALB)

x

2

f (x )=cos(x )

f (x )= ∫ e−y dy

y=0

(expcdf в MATALB)

x

−ln ( y)2

3

−|x|

f (x )=

1

∫

e

2

dy

y

√(2

f (x )=e

π) y=0

(logncdf в MATALB)

1

x

y2

∫ y e−

− x2

f (x )=

2

dy

4

√(2

f (x )=e

π) y=0

(rayldf в MATALB)

1

x

−

y2

y=−∞∫

2

f (x )=

e

dy

5

f (x )=ln (x2+1)

√

(2 π)

(normcdf в MATALB)

x

6

f (x )=x

2

f (x )= ∫ e−y dy

y=0

(expcdf в MATALB)

x

−ln ( y)2

7

f (x )=x

3

f (x )=

1

∫

e

2

dy

y

√(2

π) y=0

(logncdf в MATALB)

1

x

y2

∫ y e−

ex

f (x )=

2

dy

8

f (x )=

√(2

π) y=0

2

+1

x

(rayldf в MATALB)

1

x

−

y2

y=−∞∫

2

9

f (x )=

1

f (x )=

e

dy

√

(2 π)

x2+1

(normcdf в MATALB)

x

10

f (x )=

x2−1

f (x )= ∫ e−y dy

x2+1

y=0

(expcdf в MATALB)

1

x

e

−ln ( y)2

11

f (x )=

x2−1

f (x )=

∫

2

dy

y

√(2 π) y=0

ex

(logncdf в MATALB)

1

x

y2

∫ y e−

12

f (x )=2

−x

f (x )=

2

dy

√(2 π) y=0

(rayldf в MATALB)

1

x

e−

y2

−x2

f (x )=

∫

2

dy

13

f (x )=2

√(2

π) y=−∞

(normcdf в MATALB)

x

14

f (x )=ln (|x|+1)

f (x )= ∫ e−y dy

y=0

(expcdf в MATALB)

1

x

− ln (y )2

f (x)=

y∫=0

e

2

dy

15

f (x )=sin( x) x

√

y

(2

π)

(logncdf в MATALB)