7.2NatLog&amp;ExpSolutions

# 7.2NatLog&amp;ExpSolutions - MthSc 108-Learning...

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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 ...
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