7.2NatLog&ExpSolutions

7.2NatLog&ExpSolutions - MthSc 108-Learning...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MthSc 108-Learning Activity Solutions Section 7.2: The Natural Logarithmic and Exponential Functions 1. Let u = π e − t . Then du = −π e − t dt and dy 2 = e − t sec 2 (π e − t ), y ( ln 4 ) = π dt du = e − t dt . −π ∫e −t sec 2 (π e − t )dt = − sec π∫ 2 . 1 2 udu = − 1 π tan u + C = − 1 π tan(π e − t ) + C Now find C so that y ( ln 4 ) = y ( ln 4 ) = 2 2 =− 1 π π π tan(π e − t ) + C π 2 =− =− 1 π 1 tan (π e − ln 4 ) + C and thus y = − 1 π tan + C π π 4 2 1 = − (1) + C π tan(π e − t ) + 3 π . π π C= 3 π 1 du = e 2 x dx . 2 2. Let u = 1 + e 2 x . Then du = 2e 2 x dx and then 11 1 e2 x 1 2x ∫ 1 + e2 x dx = 2 ∫ u du = 2 ln u + C = 2 ln 1 + e + C e2 x 1 2x ∫ 1 + e2 x dx = 2 ln 1 + e + C 3. For the third term of the integrand, let u = − x . Then du = − dx and − du = dx . e2 x 2e x 1 e2 x + 2e x + 1 dx = ∫ x + x + x dx = ∫ ex e e e e2 x + 2e x + 1 x −x ∫ ex dx = e + 2 x − e + C ∫ (e x + 2 + e− x dx = e x + 2 x − e− x + C ) 4. Let u = 1 + 2 x = 1 + ( 2 x ) du = 1/ 2 . Note that 2 x = u − 1 . Then 1 1 −1/ 2 dx and ( 2 x ) ⋅ 2dx = 2 2x 1 2x dx 2 xdu = dx or (u − 1)du = dx . ∫ 1+ = ∫ u −1 du u 1 = ∫ 1 − du u = u − ln u + C1 = 1 + 2 x − ln 1 + 2 x + C1 = 2 x − ln 1 + 2 x + C where C = C1 + 1 ( ) ∫ 1+ 1 2x dx = 2 x − ln 1 + 2 x + C 5. Let u = ln x . Then du = 16 1 dx . When x = 2, u = ln 2 and when x = 16, u = 4 ln 2 . x ∫x 2 dx ln x du u 4 ln 2 ln 2 4 ln 2 = ln 2 ∫ =2 u = 4 ln 2 − 2 ln 2 = 2 ln 2 16 ∫x 2 dx dx = 2 ln2 ln x ...
View Full Document

Ask a homework question - tutors are online