Unformatted text preview: 9. x sin1 x d x U = sin1 x dx dU = 1  x2 1 1 = x 2 sin1 x  2 2 = dV = x dx x2 V = 2 2 dx x Let x = sin 1  x2 d x = cos d 1 1 2 1 x sin x  sin2 d 2 2 1 2 1 1 = x sin x  (  sin cos ) + C 2 4 1 2 1 1 = x  sin1 x + x 1  x 2 + C. 2 4 4 3. x 2 cos x d x U = x2 dU = 2x d x d V = cos x d x sin x V = 2 x 2 sin x  x sin x d x = U =x d V = sin x d x cos x dU = d x V = 2 1 x 2 sin x x cos x  +  cos x d x = 1 2 2 = x 2 sin x + 2 x cos x  3 sin x + C. 2. (x + 3)e2x d x U = x + 3 d V = e2x d x 1 dU = d x V = 2 e2x 1 1 e2x d x = (x + 3)e2x  2 2 1 1 = (x + 3)e2x  e2x + C. 2 4 /4 20. Area A = 2 =2 0 /4 0 (cos2 x  sin2 x) d x
/4 cos(2x) d x = sin(2x) 0 y = 1 sq. units. y=cos2 x y=sin2 x A x 4 Fig. 720 5/4 17. Area of R = /4 (sin x  cos x) d x
5/4 = (cos x + sin x) /4 = 2 + 2 = 2 2 sq. units.
y y=sin x R /4 y=cos x 5/4 x Fig. 717 11. For intersections: 1 5  2x =y= . x 2 Thus 2x 2  5x + 2 = 0, i.e., (2x  1)(x  2) = 0. The graphs intersect at x = 1/2 and x = 2. Thus 2 1 5  2x  dx Area of R = 2 x 1/2 =
2 5x x2   ln x 2 2 1/2 15  2 ln 2 sq. units. = 8
1 2 ,2 y 2x+2y=5 R y=1/x 1 2, 2 x Fig. 711 6. For intersections: 7 + y = 2y 2  y + 3 2y 2  2y  4 = 0 2(y  2)(y + 1) = 0 i.e., y = 1 or 2.
2 Area of R =
2 1 [(7 + y)  (2y 2  y + 3)] dy =2 1 (2 + y  y 2 ) dy 1 2 1 3 y  y 2 3
2 1 = 2 2y + = 9 sq. units.
y (9,2) x=2y 2 y+3 R xy=7 x (6,1) Fig. 76 2 3. Area of R = 2 0 (8  2x 2 ) d x
2 0 4 = 16x  x 3 3 = 64 sq. units. 3
y y=3x 2 2 R 2 x y=x 2 5 Fig. 73 1 2. Area of R = = 0 ( x  x 2) d x
1 0 2 3/2 1 3 x  x 3 3 = 2 1 1  = sq. units. 3 3 3
y y= x R y=x 2 (1,1) x Fig. 72 2 47. Area R = 0 x4
4 0 x dx + 16 1 = 2 Let u = x 2 du = 2x d x du 1 u = tan1 2 + 16 u 8 4
y 4 0 = sq. units. 32 y= x x 4 +16 R x Fig. 647 12. ln t dt t = Let u = ln t dt du = t 1 2 1 u du = u + C = (ln t)2 + C. 2 2 8. = x 2 2x 3 +1 1 3 3 2x +1 + C. = 3 ln 2 Let u = x 3 + 1 du = 3x 2 d x 1 2u +C 2u du = 3 ln 2 dx ...
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 Fall '06
 GROSS
 Calculus

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