finding the indefinite integral 1+e^x+1/x?

Question

1 + e^x + 1/x; f(1)=3 + e

I am really stumped on this one. Please if you have any hints I would really appriciate it .

Answer 1

Integrate the individual terms to get that integral, f(x) = x + e^x + ln(x) + C.

f(1) = 3 + e

1 + e + c = 3 + e --> C = 2

f(x) = x + e^x + ln(x) + 2

Answer 2

Since integration is commutative over addition just integrate each term indivdually.

1 --> x + constant
e^x --> e^x + constant
1/x = ln x + constant

Combine all the constants into only one to get
F(x) = x + e^x + ln(x) + k.

Since f(1) = 3 + e, plug 1 into F.
F(1) = 1 + e^1 + ln(1) + k = 1 + e + 0 + k = e + k + 1.
Set e + k + 1 = e + 3 (the known answer) to find that k = 2.

Thus F(x) = x + e^x + ln(x) + 2, for x > 0.

Answer 3

Simply separate into 3 integrals:
∫ 1 dx + ∫ e^x dx + ∫ 1/x dx
x + e^x + ln(x)
You just need to know that the integral of e^x is e^x and the integral of 1/x is ln(x).

x + e^x + ln(x) + C

Now you need to solve for C
f(1) = 3 + e
1 + e^1 + ln(1) + C = 3 + e
1 + e + 0 + C = 3 + e
1 + e + C = 3 + e
C = 2
so:

f(x) = x+ e^x + ln(x) + 2

Answer 4

Find the integral of each part.
Integral of 1 with respect to x is = x
Integral of e^x with respect to x is = e^x
Integral of e^x with respect to x is = ln x

x + e^x + ln x + C

Now plug in 1 for X.

1 + e + 0 + C = 3 + e

C = 2
x + e^x + ln x + 2

Answer 5

The integral a sum of expressions is the sum of the integrals of the expressions.

In this case, you get: int(1 dx) + int(e^x dx) + int (1/x dx) =

x + e^x + ln(x) + c

So, now substitute in 1 for x, and you get:

1 + e + 0 + c = 3 + e

So, c = 2

So, the total is:

x + e^x + ln(x) + 2

Answer 6

integrating each term

f(x) = x + e^x + ln x + C
where C is constant of integration
put x = to get f(1) = 1 + e + C = 3+ e so C = 2

so f(x) = x + e^x + ln x + 2

Answer 7

Answers to H_Book Questions:
Appendices: Integrals of Motion: Answers Page 1 Answers Page 2


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uestion:

Utilizing the general expression for the time-derivative of any unit vector, derive the above expression [VI.I.7] for the acceleration (Dv) from the definition of v.



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nswer:
Realizing that the general (curvilinear coordinate) form of the velocity is


v = e1 (h1Dx1) + e2 (h2Dx2) + e3 (h3Dx3),
we can, quite generally, write the time-derivative of v as,


Dv = e1 [ d(h1Dx1)/ dt ] + e2 [ d(h2Dx2)/ dt ] + e3 [ d(h3Dx3)/ dt ]
+ (h1Dx1) [ de1/dt ] + (h2Dx2) [ de2/dt ] + (h3Dx3) [ de3/dt ].


Drawing upon the following table to provide the general expression for the time-derivative of any unit vector,


Evaluating: (¶xjei)
¶x1 ¶x2 ¶x3
e1 - e2 f3 - e3 f5 e2 f1 e3 f2
e2 e1 f3 - e3 f6 - e1 f1 e3 f4
e3 e1 f5 e2 f6 - e1 f2 - e2 f4



where: f1 º (1/h1) ¶x1h2;
f2 º (1/h1) ¶x1h3;
f3 º (1/h2) ¶x2h1;
f4 º (1/h2) ¶x2h3;
f5 º (1/h3) ¶x3h1;
f6 º (1/h3) ¶x3h2.

we derive,


Dv = e1 D(h1Dx1) + e2 D(h2Dx2) + e3 D(h3Dx3)
+ (h1Dx1) [e2 ( f1Dx2 - f3Dx1 ) + e3 ( f2Dx3 - f5Dx1 )]
+ (h2Dx2) [e1 ( f3Dx1 - f1Dx2 ) + e3 ( f4Dx3 - f6Dx2 )]
+ (h3Dx3) [e1 ( f5Dx1 - f2Dx3 ) + e2 ( f6Dx2 - f4Dx3 )].

Finally, grouping terms with like vector components,


Dv = e1 { D(h1Dx1) + (h2Dx2)( f3Dx1 - f1Dx2 ) + (h3Dx3)( f5Dx1 - f2Dx3 ) } +
e2 { D(h2Dx2) + (h1Dx1)( f1Dx2 - f3Dx1 ) + (h3Dx3)( f6Dx2 - f4Dx3 ) } +
e3 { D(h3Dx3) + (h1Dx1)( f2Dx3 - f5Dx1 ) + (h2Dx2)( f4Dx3 - f6Dx2 ) },

or,


Dv = e1 { D(h1Dx1) - (h2Dx2) A - (h3Dx3) B } +
e2 { D(h2Dx2) + (h1Dx1) A - (h3Dx3) C } +
e3 { D(h3Dx3) + (h1Dx1) B + (h2Dx2) C }.


where:
A º f1Dx2 - f3Dx1 = [Dx2] (1/h1) ¶x1h2 - [Dx1] (1/h2) ¶x2h1
B º f2Dx3 - f5Dx1 = [Dx3] (1/h1) ¶x1h3 - [Dx1] (1/h3) ¶x3h1
C º f4Dx3 - f6Dx2 = [Dx3] (1/h2) ¶x2h3 - [Dx2] (1/h3) ¶x3h2


Q.E.D.


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uestion:

Derive the expressions for the magnitude | J |, and the two components Jz and Jx of the specific angular momentum vector J º r ´ v in terms of the system of spherical coordinates ( x1, x2, x3 ) = ( r, q, j ).



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nswer:
In spherical coordinates, the specific angular momentum vector takes the form:


J º r ´ v
= [ er r ] ´ [ er ( Dr ) + eq ( r Dq ) + ej ( r sinq Dj ) ]
= ej ( r2 Dq ) - eq ( r2 sinq Dj ) .

Hence,


| J | º ( J × J )1/2
= [ ( r2 Dq )2 + ( r2 sinq Dj )2 ]1/2 .

Also, we know from the direction cosines in spherical coordinates that the unit vectors eq and ej expressed in Cartesian Coordinates are:


eq = i cosq cosj + j cosq sinj - k sinq ,
ej = - i sinj + j cosj .

Hence,


J = ( r2 Dq ) [ - i sinj + j cosj ]
- ( r2 sinq Dj ) [ i cosq cosj + j cosq sinj - k sinq ]
= - i [ r2 sinj Dq + r2 sinq cosq cosj Dj ]
+ j [ r2 cosj Dq - r2 sinq cosq sinj Dj ]
+ k [ ( r sinq )2 Dj ] .