Calculus Quiz 1 ∑

Calculus Quiz 1

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Id #:

1. (5 points) Evaluate Z _π/6

0

e^xcos x dx

Solution: Let I= Z _π/6

0

e^xcos x dx=e^xsin x|^π₀^/6− Z _π/6

0

e^xsin x dx

=e^xsin x|^π₀^/6+e^xcos x|^π₀^/6− Z _π/6

0

e^xcos x dx=¡1 2+

√3

2

¢e^π/6−e−I

Therefore, I=¡1 4+

√3

4

¢e^π/6−e 2.

2. (5 points) Find

Z x²+1 x²+3x+2dx Solution:

Z x²+1

x²+3x+2dx= Z ¡

1+ −3x−1 x²+3x+2

¢dx=x+

Z ¡ A

x+1+ B x+2

¢dx

where A,B satisfy that−3x−1=A(x+2) +B(x+1)for all x∈R.

By setting x=−1 and x=−2,we get A=2 and B=−5,respectively.

Therefore,

Z x²+1

x²+3x+2dx=x+2ln(x+1)−5ln(x+2).

3. (5 points) Show that Z _∞

1

√ 1

1+x²dx is divergent.

Solution: Since lim

x→∞

√ 1 1+x²

1 x

=1,and Z _∞

1

1

xdx=lnx|^∞₁ =∞, Z _∞

1

√ 1

1+x²dx is divergent by the limit comparison test.

4. (5 points) Use the trapezoidal rule to approximate the integral Z ₃

1

x³dx with n =5. Compare your approximation with the exact value.

Solution: Note that x_i=1+2i

5 =5+2i

5 for i=0,1, . . . ,5.

Hence, T₅(x³) =1 5

¡1+27+2

∑

(5+2i)³ 125

¢=20+ 8 25. Since the exact value is

Z ₃

1

x³dx=x⁴

4 |³₁=20,the error is 8 25.

5. (5 points) Find the Taylor series of f(x) =tan⁻¹x at x=0.

Solution: Note that d tan⁻¹x

dx = 1

1+x²,and 1

1−u=

∑

k=0

u^k.We get that 1

1+u =

∑

k=0

(−1)^ku^k, which implies that 1

1+x² =

∑

k=0

(−1)^kx^2k.Hence, tan⁻¹x=

Z 1

1+x²dx=

∑

k=0

(−1)^kx^2k+1 2k+1 .

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