Math 102- Solved Problems (25 Problems)

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Dr. Ibrahim Al-Rasasi

Math 102- Solved Problems

(25 Problems)

1. Write the sum

∑

= + − 20 1 2 3 ) 5 sin( n n n

in a sigma notation that starts with 3 rather than 1. Solution: Add 2 to both sides of n=1 to get n+2=3. Let j= n+2. Then

2 − = j

n . Further, n=1⇒ j =3 and n=20⇒ j=22. Thus we have

∑

= = − − = + − 20 1 22 3 2 1 ) 7 sin( 3 2 ) 5 sin( n j j j n n . 2. Evaluate the limit

∑

= +∞ → n i n n i n 1 ) sin( 1

lim . Solution: We compare with the

formula

∑

∫

= +∞ → Δ = n i b a i n f x x f x dx 1 * . ) ( ) (

lim We observe that

n x= 1 Δ and ; * n i

x_i = that is, x is chosen to be the right endpoint of each subinterval. Since _i* x i a xi = + Δ * , then n i a n i 1 +

= and hence a=0.Also, since , n a b x= − Δ then n b n 0 1 ₌ −

and hence b=1. Further, f(x)=sinx. So the given limit equals

∫

1

0

sin x dx=1−cos1.

3. Find the critical points of the function

∫

− = x x t dt e x F 2 2 2 2 . )

( Solution: We find the derivative: F'(x)=e(x2−2x)2(2x−2)=0⇒ x=1. So the function has only one critical point: x=1.

4. Find

∫

cos2 x dx. Solution: Using the identity cos(2x)=2cos2x−1, we have

∫

cos2 x =

∫

+ = + sin(2 )+ . 4 1 2 1 2 ) 2 cos( 1 C x x dx x dx 5. Find . 2 2 dx x x

∫

₊ ₋ Solution: Multiplying the numerator and denominator by the conjugate x+ 2+ x , the integral becomes

∫

( x+2+ x)dx=

. ] ) 2 [( 3 2 3/2 3/2 C x x+ + + 6. Find . 1 2 2 dx x x

∫

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

+ Solution: First divide

2

x by x+1 (use long division) to

get 1 1 1 1 2 + + − = + x x x x . Squaring, we get + + − + − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + 1 2 2 ) 1 ( 1 2 2 2 x x x x x 2 ) 1 ( 1 +

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divide first: 1 4 2 1 2 2 + − = + − x x x . Thus, . 1 1 | 1 | ln 4 2 3 ) 1 ( 1 3 2 2 C x x x x dx x x ₊ + − + − + − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +

∫

7. Find

∫

xtan(3+x3/2)dx.Solution: Let u=3+x3/2. Then . 2 3 1/2 dx x du = We get

∫

x +x dx=

∫

tanu 3 2 ) 3 tan( 3/2 du= u +C = lnsec(3+x ) +C 3 2 sec ln 3 2 3/2 . 8. Consider the function

x x

f( )= over the interval [1, 6]. Find the values of 1 c that satisfy the Mean Value Theorem for Integrals. Solution: since f is continuous on the interval [1, 6], then, by MVTI, there is a number c in [1, 6]

such that

∫

= − 6 1 ). 1 6 )( ( ) (x dx f c f This gives c 5 6 ln = and hence . 6 ln 5 = c Note that [1,6] 6 ln5 ∈ since ln6 6. 5 1 6 ln 6 5 6 ln 6 6< e5 < 6 ⇒ < < ⇒ < <

9. Find

∫

xln(x2 +1)dx. Solution: Let .y= x2 +1 Then dy=(2x)dx.Thus we get

∫

x x + dx=

∫

lny 2 1 ) 1 ln( 2 dy= y y−y +C= y[lny−1]+C 2 1 ] ln [ 2 1 = ( 1)[ln( 1) 1] . 2 1 2 2 C x x + + − + 10. Find

∫

csc6(2x)dx. Solution:

∫

csc6(2x)dx=

∫

csc4(2x)csc2(2x)dx

∫

+ =

∫

+ +

= (1 cot2(2x))2csc2(2x)dx (1 2cot2(2x) cot4(2x))csc2(2x)dx. let ).

2 cot( x

u= Then du =−2csc2(2x)dx. The last integral equals ) 2 ( cot 3 1 ) 2 cot( 2 1 ) 5 1 3 2 ( 2 1 ) 2 1 ( 2 1 2 4 3 5 3 x x C u u u du u u + = − + + + = − − + −

∫

. ) 2 ( cot 10 1 5 C x + − 11. Find .(sin ) 1 1 11 dx x

∫

−

Solution: Here we do not need to use the technique of section 7.2. Since the interval is symmetric and the integrand is an odd function (check), then

∫

− = 1 1 11 . 0 ) (sin x dx

12. Find

∫

1+9x2dx. Solution: Let I denote the given integral. First let y=3x. Then dy=3dx. We get =

∫

1+ . 3 1 2 dy y I Let . 2 2 , tanθ −π <θ <π = y Then θ θ d

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integration by parts: Let u=secθ, dv=(sec2θ)dθ. Then ,

) tan (secθ θ dθ

du= v=tanθ. This gives

∫

= − +

−

= [sec tan sec ( ) (sec ) ]

3 1 ] ) tan (sec tan [sec 3 1 θ θ θ 2θ θ θ θ 3 θ θ θ θ d d d I |]. tan sec | ln 3 tan [sec 3 1 _θ _θ ₋ ₊ _θ ₊ _θ

= I Solving for ,I we get

. |] tan sec | ln tan [sec 6 1 C I = θ θ + θ + θ + In terms of x we get , . |] 3 9 1 | ln 9 1 3 [ 6 1 2 2 C x x x x I = + + + + + 13. Find

∫

₋ −2 . 1 dx e

ex x Solution: Multiply the integrand by x x e e to get

∫

₋ . 2 2 dx e e x x

Let u=ex. Then du=exdx. The last integral becomes . 2 2 ln 2 2 1 2 2 ln 2 2 1 2 1 2 C e e C u u du u x x + + − = + + − = −

∫

14. Find the limit of the sequence } . 2 ) 5 cos( { ₁ 3 ∞ + = + n n n

Solution: We use the Squeezing Theorem. Since −1≤cos(5n)≤1, then .

2 1 2 ) 5 cos( 2 1 3 3 3 n n n n ≤ + ≤ + + − Since 0 2 1 lim 3 = + − +∞ → n n and 0, 2 1 lim 3 = + +∞ → n n then 0. 2 ) 5 cos( lim 3 = + +∞ → n n n

15. Determine whether the series

∑

+∞ =1 − 2 1 4 1 n n converges or diverges. If it converges , find its sum. Solution: By definition, we need to find lim_n_→_+∞ S_n.

We have ⎟= − + − + ⎠ ⎞ ⎜ ⎝ ⎛ + − − = − =

∑

= = ) 5 1 3 1 ( ) 3 1 1 1 [( 2 1 1 2 1 1 2 1 2 1 1 4 1 1 1 2 n k n k n k k k S ). 1 2 1 1 ( 2 1 )] 1 2 1 1 2 1 ( ... ) 7 1 5 1 ( + − = + − − + + − n n

n This implies that

. 2 1

lim_n_→_+∞ S_n = We conclude that the series converges and its sum is . 2 1

16. Find all values of x for which the series

∑

+∞ = − 0 n xn

e converges and find the sum of the series for theses values of x. Solution: The series

∑

+∞ = − − − ₌ ₊ ₊ ₊ 0 2 ... ) ( 1 ) ( n x x n x e e

e is a geometric series with a=1 and r =e−x. Thus, it converges if r <1, i.e., e−x <1⇒e−x <1⇒ex >1⇒ x>0. The series converges if and only if x>0 and the sum in this case is

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∑

∞ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 1 2 1 1 n n

n converges or diverges. Solution: We use the Divergence Test: since lim 1 1 2 0,

2 ≠ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − +∞ → e n n

n then the given

series diverges.

∑

+∞

=2 ln

1

n n n

converges or diverges. Solution: We use the Integral Test:

x x x f ln 1 )

( = is positive, continuous and decreasing ) 0 ) ln ( 2 ) ln 2 ) ((ln ) ( ' ( 2 2 / 1 < + − = − x x x x x

f on [2,+∞ . So we can apply the Integral ) Test:

∫

+∞ +∞ → +∞ → +∞ → = = − =+∞ = 2 2 2 lim 2 ln 2 ln2 . | ln 2 lim ln 1 lim ln 1 t t t t t dx x t x x dx x x

Since the improper integral diverges, then the given series diverges. 19. Determine whether the series

∑

+∞

=1

1 n n n

converges or diverges. Solution: The series can be rewritten as

∑

+∞ = +∞ = = 1 1 2 / 3 . 1 1 n n n n n

Thus the series is a p-series with 1

2 3 > =

p and hence it is convergent. 20. Determine whether the series

∑

+∞ = − 1 3 1 ) / 1 ( tan n n n

converges or diverges. Solution: We will use the Comparison Test. Since tan−1(1/n)≤π/2 for all n≥1, then

3 3 1 2 ) / 1 ( tan n n n _≤ π −

for all n≥1. Since

∑

+∞

=12 3

n n

π

converges (as it is a constant times a p-series with p=3>1 ), then the series

∑

+∞ = − 1 3 1 ) / 1 ( tan n n n converges. 21. Determine whether the series

∑

+∞ = + + 14 5 3 2 n n n

converges or diverges. Solution: We use the Limit Comparison Test. Choose .

5 3 5 3 n n n n b ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = Then _→_+∞ = n n n b a lim . 0 1 1 0 1 0 1 5 1 4 1 3 1 2 lim ) 1 5 1 4 ( 15 ) 1 3 1 2 ( 15 lim 15 3 4 15 5 2 lim 3 5 5 4 3 2 lim = > + + = + ⋅ + ⋅ = + ⋅ + ⋅ = + ⋅ + ⋅ = ⋅ + + n n n n n n n n n n n n n n

Since the series

∑

∞ + = ∞ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 1 1 5 3 n n n n

b converges (a geometric series with 1

3 < =

r ), then the series

∑

+∞ ₂₊₃n

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∑

+∞ =1 ) cos( n n nπ

converges or diverges. Solution: Note first that cos(nπ)=(−1)n. Thus, the series

∑

+∞ = +∞ = − = 1 1 ) 1 ( ) cos( n n n n n nπ is an alternating series. So we use the AST: 1 ;

n a_n = (i) lim _→_+∞ 1 =0, n n and (ii) 0 2 1 ) ( ' 1 ) ( 3 < − = ⇒ = x x f x x

f for x≥1, and hence

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ n 1 is decreasing. We conclude that the series

∑

+∞ =1 ) cos( n n nπ

converges by the AST. 23. Determine whether the series

∑

+∞ = + − 1 3 . 0 1 ) 1 ( n n n is absolutely convergent, conditionally convergent, or divergent. Solution: We check each possibility:

a. Absolute Convergence: Since the series

∑

+∞ = +∞ = + = − 1 3 . 0 1 3 . 0 1 1 ) 1 ( n n n n n diverges

(a p-series with p=0.3<1), then the given series is not absolutely convergent.

b. Conditional Convergence: We have to check the two conditions of conditional convergence: i.

∑

+∞ = + − 1 3 . 0 1 ) 1 ( n n

n diverges: This was shown in (a). ii.

∑

+∞ = + − 1 3 . 0 1 ) 1 ( n n

n converges: It is an alternating series and so we use the AST. We leave it to the reader to check that the series converges. Conclusion: The given series converges conditionally.

∑

+∞ = + + − 1 1 )! 2 3 ( ) 3 ( n n

n converges or diverges. Solution: We use the Ratio Test:

)! 2 3 ( ) 3 ( 1 + − = + n a n n , )! 5 3 ( ) 3 ( )! 2 ) 1 ( 3 ( ) 3 ( 2 2 1 ₊ − = + + − = + + + n n a n n n . )! 2 3 ( ) 3 3 )( 4 3 )( 5 3 ( )! 2 3 ( ) 3 ( lim ) 3 ( )! 2 3 ( )! 5 3 ( ) 3 ( lim lim ₁ 2 1 + ⋅ + + + + ⋅ − = − + ⋅ + − = = + + ₊ +∞ → n n n n n n n a a L _n n n n n . 0 ) 3 3 )( 4 3 )( 5 3 ( 3 lim = + + + = _→_+∞ n n n

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∑

∞ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 1 2 3 n n n n

converges or diverges. Solution:

We use the Root Test: ⎟ =

⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = = _→_+∞ n n n n n n n n n a

L lim lim 3 lim 1 3

2

3