Assignment1_Solns

Assignment1_Solns - Section 5.3 54 Evaluate the integral...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

Assignment1_Solns - Section 5.3 54 Evaluate the integral...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online