Unformatted text preview: nanni (arn437) – Assignment 14 – guntel – (54940)
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001 since both f and g are even functions. To
evaluate this integral we must ﬁrst use partial
fractions:
5
5
=
2
9−x
(3 − x)(3 + x) 10.0 points The bounded region enclosed by the graphs 5
1
1
.
+
6 3−x 3+x = of 5
,
9 − x2
is the one shaded in
f ( x) = g ( x) = 1 , Thus the integral becomes
2 2
graph of g 1
1
5
+
6 3−x 3+x 1−
0 = 2 x− 1 1 5
3+x
ln
6
3−x dx
2
0 . Consequently,
area = 4 −
graph of f
−2 2 Find the area of this region. 002 10.0 points Evaluate the integral
2 I= 5
1. area = 2 − ln 4
6 5
ln 5 .
3 0 2. area = 4 + 5
ln 4
3
5
ln 5 correct
3 4x − 8
dx .
− 2x − 3 1. I = 4 ln 4 3. area = 4 − x2 2. I = −4 ln 4
3. I = −2 ln 4 5
4. area = 2 + ln 5
6 4. I = 2 ln 3 correct 5
5. area = 2 − ln 5
3 5. I = −2 ln 3 5
6. area = 4 + ln 4
6 6. I = 4 ln 3 Explanation:
The area of the shaded region is given by
the deﬁnite integral Explanation:
After factorization
x2 − 2x − 3 = (x + 1)(x − 3) . 2 (g (x) − f (x)) dx
But then by partial fractions, −2
2 =2 1−
0 5
dx
9 − x2 3
1
4x − 8
=
+
.
x2 − 2 x − 3
x+1 x−3 nanni (arn437) – Assignment 14 – guntel – (54940)
Thus Now
2
0 3
dx =
x+1 2 3 ln (x + 1) = 3 ln 3 , 0 while
2
0 2 I= 1− 3
1
+
x+1 x−2 dx . Now
1
dx =
x−3 2 ln (x − 3) 0 3
dx = 3 ln x + 1 + C1 ,
x+1 = − ln 3 . Consequently, while
I = 2 ln 3 . 003 10.0 points Determine the integral
x2 − 3 x + 5
dx .
x2 − x − 2 I= 1. I = x + ln (x − 1)3
+C
x+2 2. I = x − ln (x + 1)3 (x − 2) + C
3. I = x − ln (x − 1)3
+C
x+2 1
dx = ln x − 2 + C2 .
x−2
Consequently,
(x + 1)3
+C .
I = x − ln
x−2 keywords: division, partial
004 10.0 points
What date is Exam 2?
1. November 26
2. October 19 4. I = x + ln (x − 1)3 (x + 2) + C
3. November 2 correct
5. I = x + ln 1)3 (x +
x−2 +C (x + 1)3
+ C correct
6. I = x − ln
x−2
Explanation:
By division, 5. May 18
6. October 29
Explanation:
See syllabus. x2 − 3 x + 5
x2 − x − 2
= 4. December 3 (x2 − x − 2) − 2x + 7
x2 − x − 2
= 1− 2x − 7
.
−x−2 x2 But by partial fractions,
2x − 7
3
1
=
−
.
x2 − x − 2
x+1 x−2 ...
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 Fall '10
 Gualdini
 Calculus, Derivative, 1982

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