Assignment1_Solns

# Assignment1_Solns - Section 5.3 54 Evaluate the integral...

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Section 5.3 54) Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch. Z 2 π π/ 6 cos x dx. Solution: Z 2 π π/ 6 cos x dx = [sin x ] 2 π π/ 6 = 0 - 1 2 = - 1 2 . It is a sum of the two areas with + and subtracts the area with a - symbol. 0 1 2 3 4 5 6 ï 1 ï 0.8 ï 0.6 ï 0.4 ï 0.2 0 0.2 0.4 0.6 0.8 1 cos x + ï + 56) Find the derivative of the function. g ( x ) = Z 1+2 x 1 - 2 x t sin t dt. Solution First we separate the integral into two parts and in the second term we exchange the order of integration, Z 1+2 x 1 - 2 x t sin t dt = Z 0 1 - 2 x t sin t dt + Z 1+2 x 0 t sin t dt, = - Z 1 - 2 x 0 t sin t dt + Z 1+2 x 0 t sin t dt. 1

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Next, we compute the derivatives by using the chain rule. In the ﬁrst we make a substitu- tion of u = 1 - 2 x and in the second we choose u = 1 + 2 x . The result is, d dx Z 1+2 x 1 - 2 x t sin t dt = - d dx Z 1 - 2 x 0 t sin t dt + d dx Z 1+2 x 0 t sin t dt, = - ± d dx (1 - 2 x ) ² (1 - 2 x ) sin(1 - 2 x ) + ± d dx (1 + 2 x ) ² (1 + 2 x ) sin(1 + 2 x ) , = 2(1 - 2 x ) sin(1 - 2 x ) + 2(1 + 2 x ) sin(1 + 2 x ) , 60) If f ( x ) = R x 0 (1 - t 2 ) e
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Assignment1_Solns - Section 5.3 54 Evaluate the integral...

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