{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 2 - Solution set for Math 155 of Assignment 2 Prob 1 y= 0...

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

Solution set for Math 155, of Assignment 2 Prob. 1 y = x 0 ( 1 + u 4 ) du . Hence dy dx = d dx x 0 ( 1 + u 4 ) du . So, from Fundamental Theorem of Calculus as the function 1 + u 4 is continuous be- tween 0 and x dy dx = 1 + x 4 . Prob. 2 y = 2 x - 1 0 ( t 2 - 1 ) dt . Here the function f ( x ) = x 2 - 1 is continuous for all values of x, especially for x > 0 . Hence we can apply Leibniz’s rule. h ( x ) = 2 x - 1 , g ( x ) = 0 h ( x ) = 2 , g ( x ) = 0 Hence applying equation (1) dy dx = ( 1 + ( 2 x - 1 ) 2 ) 2 ; Prob. 3 y = ln x 3 ( e - t ) dt , x > 0 . Here the function f ( x ) = e - ln x is continuous for all values x > 0 . Hence we can apply Leibniz’s rule. h ( x ) = ln x , g ( x ) = 3 h ( x ) = 1 x , g ( x ) = 0 Hence applying equation (1) dy dx = ( e - ln x )( 1 x ) ; 1

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

View Full Document
Prob. 4 y = x - x ( tan u ) du , 0 < x < 4 . Here the function f ( x ) = tan xis continuous for all values 0 < x < 4 . Hence we can apply Leibniz’s rule. g ( x ) = - x , h ( x ) = x g ( x ) = - 1 , h ( x ) = 1 Hence applying equation (1) dy dx = ( tan x ) + ( tan ( - x )) = 0 ; Prob. 5 y = ( x 5 2 + 2 x 3 - 1 ) dx We can write the above integral as y = ( x 5 2 ) dx + 2 x 3 dx - 1 dx Thus computing the above integral, we get y = x 6 12 + x 4 2 - x + C where C is a constant.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

2 - Solution set for Math 155 of Assignment 2 Prob 1 y= 0...

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

View Full Document
Ask a homework question - tutors are online