(1) f(4x³-9x² + 7x + 3)ex. dx = A) e *[4x³ + 3x² + 13x + 16) + C B) -e [4x³ + 3x² + 13x + 16] + C C) ex (4x³-9x² + 7x+3)-e-*(12x²-18x-7)+C D) -e* (4x³-9x² + 7x-3) + e-x(12x² + 18x + 7) + C B) 2T A)7 sec 0 -2m (2) The value of fo sin(5x) cos(3x).dx= A) 1 .dx we substitute x = (3) To evaluate f √x²-49 X (4) The value of f sec tan x A)- tan x 4 + 6 C) + x + x tan³ x dx is +C B): sec x + (5) The form of the partial fraction decomposition of f(x) B C С + x² +25 x-2 B Cx+F x-2 x² +25 x² +25 (6) Evaluate the following integral A) 4 Inx-21- In|x + 3) + C C) 4 Inx-31-In/x+21+ C + + D x² +25 Dx+M + +C C)- sec x - x-2 sec x 6 B) + x + C D). x f + sec x 6 B x² +25 x-2 B Cx+F + X-2 x2+25 + x-2 (7) f√²-4x dx = A) in ²₂+√²+C B) In | | 2 4x + D) In + C) π D) 0 B) 7sine C) 7tane +C = 3x+11 x²-x-6 .dx B) In/x-31-In/x + 2) + C In (x-3) D) + C In (x+2) D) 11x¹-2x³-4x² +5 x(x-2)(x²+25)² D (x²+25)² DX+M (x²+25)2 √√x²-4x 2 tan¹ x 4 x-2 √x² - 4x x-2 C) In 2 2 (8) f csc (3x) dx = A) In |csc (3x) + cot (3x)| + C C)- In csc (3x)- cot (3x)| + C B)-In (csc (3x) + cot (3x)| + C D)- In |csc (3x) + tan (3x)| + C (9) Find an appropriate choice of u and dv for integration by parts of f(In(x))².dx. A)u = (Inx)2, dv = dx B)u = lnx, dv=dx C)u = lnx, dv=lnx. dx D)u = x,dv = (lnx)².dx tan" 6 + C √x²4x1 2 D) 7cose + C +C

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Q 1 please
(1) f(4x³-9x² + 7x + 3)ex. dx =
A) e *[4x³ + 3x² + 13x + 16) + C
B) -e [4x³ + 3x² + 13x + 16] + C
C) ex (4x³-9x2 +7x+3)-e-*(12x²-18x-7)
+ C
D) -ex (4x³9x²2 +7x-3) + e-*(12x² + 18x + 7) + C
B) 2T
-2m
(2) The value of f sin(5x) cos(3x). dx= A) 1
√√x²-49
(3) To evaluate f
X
(4) The value of f sec4 x tan³ x dx is
+ C B): sec x
A)-
tan x
4
tan x
6
+ +C C)- sec x
sec x
6
+
B
x-2
C) A+ B
(5) The form of the partial fraction decomposition of f(x)
C
D
С
+
x² +25
x² +25
x²+25
Cx+F
Dx+M
Cx+F
x² +25
x² +25
x²+25
Evaluate the following integral f
C
x-2
+
+
+
+
.dx we substitute x = A)7 sec 0
(6)
A) 4 Inx-21-In/x+31+
C) 4 Inx-31-In/x+21+ C
-
B)+
D).
sec x
4 6
+ C
D)
+
(7) √ √²4x dx = A) In | ²2 +2²+ C B) In
x-2
√x²
4x
+
B
x-2
B
+
x X-2
3x+11
x²-x-6
.dx
B) In/x-31-In/x + 2) + C
In (x-3)
+ C
In (x+2)
x-2
2
D) In
+
=
+
+CD)
C) π D) 0
B) 7sine
D
(x²+25)²
DX+M
(x²+25)²
√√x²-4x
2
11x-2x³-4x² +5
x(x-2)(x²+25)²
x- 2
√√x² - 4x
tan x
4
x-2
C) In
2
2
(8) f csc (3x) dx =
A) In |csc (3x) + cot (3x)| + C
C)- In csc (3x)- cot (3x)| + C
B)-In (csc (3x) + cot (3x)| + C
D)- In |csc (3x) + tan (3x)| + C
(9) Find an appropriate choice of u and dv for integration by parts of f(In(x))².dx.
A)u = (Inx)2, dv = dx
B)u = lnx, dv=dx
C)u = lnx, dv=lnx. dx
D)u = x,dv = (lnx)².dx
+ C
C) 7tane D) 7cose
√x²
-
-
2
tan"
6
4x
+ C
+ C
Transcribed Image Text:(1) f(4x³-9x² + 7x + 3)ex. dx = A) e *[4x³ + 3x² + 13x + 16) + C B) -e [4x³ + 3x² + 13x + 16] + C C) ex (4x³-9x2 +7x+3)-e-*(12x²-18x-7) + C D) -ex (4x³9x²2 +7x-3) + e-*(12x² + 18x + 7) + C B) 2T -2m (2) The value of f sin(5x) cos(3x). dx= A) 1 √√x²-49 (3) To evaluate f X (4) The value of f sec4 x tan³ x dx is + C B): sec x A)- tan x 4 tan x 6 + +C C)- sec x sec x 6 + B x-2 C) A+ B (5) The form of the partial fraction decomposition of f(x) C D С + x² +25 x² +25 x²+25 Cx+F Dx+M Cx+F x² +25 x² +25 x²+25 Evaluate the following integral f C x-2 + + + + .dx we substitute x = A)7 sec 0 (6) A) 4 Inx-21-In/x+31+ C) 4 Inx-31-In/x+21+ C - B)+ D). sec x 4 6 + C D) + (7) √ √²4x dx = A) In | ²2 +2²+ C B) In x-2 √x² 4x + B x-2 B + x X-2 3x+11 x²-x-6 .dx B) In/x-31-In/x + 2) + C In (x-3) + C In (x+2) x-2 2 D) In + = + +CD) C) π D) 0 B) 7sine D (x²+25)² DX+M (x²+25)² √√x²-4x 2 11x-2x³-4x² +5 x(x-2)(x²+25)² x- 2 √√x² - 4x tan x 4 x-2 C) In 2 2 (8) f csc (3x) dx = A) In |csc (3x) + cot (3x)| + C C)- In csc (3x)- cot (3x)| + C B)-In (csc (3x) + cot (3x)| + C D)- In |csc (3x) + tan (3x)| + C (9) Find an appropriate choice of u and dv for integration by parts of f(In(x))².dx. A)u = (Inx)2, dv = dx B)u = lnx, dv=dx C)u = lnx, dv=lnx. dx D)u = x,dv = (lnx)².dx + C C) 7tane D) 7cose √x² - - 2 tan" 6 4x + C + C
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