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# calc 2 - = 2x/2x 4 ln(2x 4(Y 10 Y=-sin^-1(5x^2 4 Sin-1 =-1...

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1) Find the Volume of the Solid generated by revolving about y-axis, circle x^2+y^2=16, by line x=4, y=4. V=2pi integral(4,0) x. (4-Square root of (-x^2+16) dx< let U=-x^2+16. 2) use shell method, about y-axis, Y=8-X^2, y=x^2, x=0 8-X^2=X^2, 8-2X^2 V= 2piintegral (0,2) x(8-2x^2) 3) Second Quadrant 9-x^2, by the x-axis, by the y-axis, about the line x=1 R=1-(-square root of (9-y)) V= integral (0,9) pi [1+suare root (9-y)]^2 dy 4)Region in First Quadrant by the line Y=25, Y=25-X^2 and by line X=5, about Y=25 R= 25-(25-X^2)= X^2 PI integral (0,5) (x^2)^2 5) Volume about xaxis, x=7y-y^2, x=0 2 PI integral (0,7) Y(7Y-Y^2) 6) About the line Y=3, x=y+6, x=y^2 2 PI integral (0,3) (3-Y) (Y+6-Y^2) 7) About Y-axis, X=2, Y=2+(X^2/4) X= square root of (y-2/4) 2 PI integral (0,2) x(2+(x^2/4)) 8) e^(xy)+xy=5 ( implicit diff) E^(13xy) [(13)(Y)+(1) dy/dx(13x)] + (1)(y)+ dy/dx (x) =0 13Ye^(13xy) + 13 xe^(13xy)+Y+dy/dx(x) = 0 Dy/dx = -y/x 9) Y= (2x+4)^x Lny= ln(2x+4)^x = XlnX(2x+4)

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Unformatted text preview: = 2x/2x+4 ln(2x+4) (Y) 10) Y= -sin^-1(5x^2+4) Sin-1 = -1/ square root of (1-x^2) = 1/suare root of (1-(5x+4)^2) (derivative of 5x^2+4 11) –cos-1(4x+7)/5-cos-1= 1/ suare root of (1-x^2) (dx) U= 4x+7/5, du = 4/5 4/5/square root of ((1-(4x+7)^2/25) 4/5/ 1/5(sq. rt of (25-(4x+7^2) = 4/ sq. rt 25- ( 4x+7^2) 12) Y = tan-1(2x)/x Use quotient rule, let u = tan-1(2x) and v = x, ------.> (Uv – Vu)/v^2 13) 2 sinn^2x dx 2 integral sin^2(x) . dx 2 integral ½ (1- cos2x) dx Integral 1 – intergral cas2x dx Let u = 2x, dx= ½ du x-1/2 integral cos du X-1/2 sinu +c sub U then 13) integral sq. rt (16-x^2) X= 4sint , so dx= 4 cost. Dt, sq. rt (16-x^2)= 4cos t Integral 4 cost 4 cost.dt= 16 intgeral cos^2 t dt 8 integral (1+cos2t) dt= 8(1+1/2 sin2t) +c 8(t+sint cost) +c Cost= cos [sin-1(x/4)]= SQ. RT (1)-(x^2/16) = ¼ sq rt (16-x^2) = 8 sin-1 (x/4)+(x/2) sq rt (16-x^2) +c 14) integral (4x^2+x+4)/(x^2)(x-4) Ax+B/x^2+2 + C/x-4 = x^2+x+4 ....
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