Kalkulus
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The document discusses the calculation of area and volume integrals for several functions. It calculates: 1) The area under the curve y = x^2 + 2x - 3 from x = 1 to x = -3 as 5/3 units. 2) The volume of a solid obtained by rotating the region bounded by the curves y = 4 - 2x from x = 0 to x = 2 about the x-axis as 32/3π units. 3) The total area under several curves by adding the individual areas, coming to a final value of 253/127 units.
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![Kalkulus IIYandhi Rahman100511003Teknik nformatikaAgustus 2011<br />6.1<br />-3∆xy = x2 + 2x – 31YHampiri luas irisan masing-masing :∆A : ( xi2 + 2xi – 3) ∆xJumlahkanA ≈ Σ ( x2 + 2x – 3) ∆xAmbil Limit : A = -3∫1 x2 = 2x -3 = -1/3 x3 – x2 + 3x= ( -1/3 (1)2 – 12 + 3.1 – (-1/3 (-3)2 – (-3)2 + 3.-3)= -1/3 – 1 + 3= 5/3 – 9= 5/3 + 27/3 = 32/34)<br />X<br />Sumbu XXYy = 4 – 2x∆x2<br />Batas0 = 4 – 2x-4 = -2xX = 26.2<br />4)<br />∆V ≈ π (4 – 2x)2 ∆xV = π 0∫2 (16 – 16x + 4x2)dx = 16x – 8x2 + 4/3 x3 ]02 = π (16(2) – 8(2)2 + 4/3 (2)3) = π (32/3) = 32/3 π<br />XYy = 4 – 2x X = y – 4/ –2∆x4Sumbu Y<br />∆V ≈ π (y – 4/ –2)2 ∆yV = π 0∫4 (y2 – 8y + 16/4)dy = π ¼ [ 1/3 y3 – 4y2 +16y]04 = ¼ π (1/3 (4)3 – 4(4)2 +16(4)) = ¼ π (64/3) = 64/12 π = 16/3 πBatasy = 4 – 2(0) = 4<br />03-2xyy = x3 – x2 – 6x6.1<br />-2∫0 x3 - x2 – 6x= ¼ x4 – 1/3 x3 – 3x2]-20=- ( 1/4 (-2)4 – 1/3 (-2)3 – 3(-2)2= - (4 + 8/3 – 12 )= - (- 8 + 8/3)= - (- 24/3 +8/3)= 16/3Batasx ( x - x - 6 )x1 = 0( x – 3) ( x + 2 )x = 0,x = 3,x = 2- 0∫3 = x3 - x2 – 6x= ¼ x4 – 1/3 x3 – 3x2]03= - ( 1/4 (3)4 – 1/3 (3)3 – 3(3)2= - ( 81/4 – 9 – 27 )= 81/4 – 36= 81/4 – 144/4 = 63/4Tot = 16/3 + 63/4 = 64 + 189/12= 253/127)<br />6.1<br />Y8)<br />y = -x + 2<br />y = x2<br />1-2X<br />x2 = -x +2x2 + x – 2 = 0(x+2) (x-1)x = -2 x = 1<br />= -2∫1 ( -x + 2) –x2 dx= -2∫1 –x2 – x + 2= (1/3 x3 – 1/2 x2 + 2x ]-21= -1/3 (1)3 – 1/2 (1)2 + 2(1) – (-1/3 (-2)3 – 1/2 (-2)2 + 2(-2))= -1/3 – 1/2 + 2 – (8/3 - 2 - 4)= -2 – 3 /6 + 2 – 8/3 + 6= -5/6 – 8/3 + 8= -21/6 + 8= 27/6 = 4 ½ <br />6.2<br />1)<br />∆xxxyy = x2 + 102∆xyxy = x2 + 1∆V = π(x2 + 1 )2 ∆xV = 0∫2 π (x4 + 2x2 + 1) dxV = 0∫2 π (x4 + 2x2 + 1) dxV = 1/5 x5 + 2/3 x3 + x ]02 =1/5 (2)5 + 2/3 (2)3 + 2 =32/5 + 16/3 + 2 = 96+80+30/15 = 206/15 π<br />](https://image.slidesharecdn.com/yandhicalculus-111005075727-phpapp01/85/Kalkulus-1-320.jpg)
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