{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

wk3quiz_sol

# wk3quiz_sol - Name Score 10B Practice Quiz Please return...

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

Name: Score: 10B Practice Quiz Please return during Monday’s lecture. You may use whatever resources you like. 1. (5 points) Find the area between the x - axis and the function f ( x ) = (1 + ln( x )) 2 x on the interval [1 , 3]. Solution: Finding the area between the x - axis and the function on [1 , 3] is the same as the following definite integral 3 1 (1 + ln( x )) 2 x dx For this, try the substitution, w = 1 + ln( x ). This means that dw = (1 /x ) dx . Substituting everything in, we get w 2 x x dw = w 2 dw Now applying the power rule, we get w 2 dw = (1 / 3) w 3 + C Now replacing the w with x we get (1 + ln( x )) 2 x dx = (1 / 3) w 3 + C = (1 / 3)(1 + ln( x )) 3 + C Now applying the first fundamental theorem of calculus, we get 3 1 (1 + ln( x )) 2 x dx = (1 / 3)((1 + ln(3)) 3 ) - 1)

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

View Full Document
2. (5 points) Compute the following indefinite integral (1 + tan( x )) 3 cos 2 ( x ) dx Solution: This is a substitution problem. First since (1 / cos 2 ( x ) = sec 2 ( x ), we can write (1 + tan( x )) 3 cos 2 ( x ) dx = (1 + tan( x )) 3 )sec 2 ( x ) dx Now let w = 1 + tan( x ), implying that dw = sec 2 ( x ) dx . Now substituting everything in, we get w 3 sec 2 ( x )(1 / sec 2 ( x )) dw = w 3 dw = (1 / 4) w 4 + C
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}