# CalcWA2 Brandon M.docx - Brandon Moyer Calculus...

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Brandon Moyer Calculus II (MAT-232-GS001) Written Assignment 2 University ID: 0361809 Thomas Edison State University Section no.: 6.2, 6.3, 6.4, 6.6, 3.2 Answer all assigned exercises, and show all work. Each exercise is worth 4 points. Section 6.2 4. Evaluate the integral. ln x x dx u = ln ( x ) ,dv = xdx ,du = 1 x dx ,v = x 2 2 ¿ ln ( x ) ( x 2 2 ) ( x 2 2 ) 1 x dx = 1 2 x 2 ln ( x ) 1 2 xdx = 1 2 x 2 ln ( x ) 1 2 ( x 2 2 ) + C ¿ 1 2 x 2 ln ( x ) 1 4 x 2 + C 6. Evaluate the integral. ln x dx x u = ln ( x ) ,du = 1 x dx ¿ udu = ln ( x ) ( 1 x ) dx ¿ ln 2 x 2 + C 18. Evaluate the integral. 2 sin x x dx u = x 2 ,du = 2 xd x →xdx = 1 2 du xsin ( x 2 ) dx = 1 2 sin udu = 1 2 cos u + C ¿ 1 2 cos ( x 2 ) + C WA 2, p. 1
Brandon Moyer Calculus II (MAT-232-GS001) Written Assignment 2 22. Evaluate the integral. 1 2 3 0 x x e dx u = x 2 ,dv = e 3 x dx,du = 2 xdx,v = e 3 x 3 ¿ x 2 e 3 x 3 | 0 1 0 1 e 3 x 3 2 xdx = 1 3 ( e 3 0 ) 2 3 0 1 x e 3 x dx u again …u = x ,dv = e 3 x dx ,dv = dx ,v = e 3 x 3 xe 3 x dx = ¿ e 3 3 2 3 ( x e 3 x 3 | 0 1 0 1 e 3 x 3 dx ) = e 3 3 2 3 ( ( e 3 3 ) 0 1 e 3 x 3 dx ) = ¿ e 3 3 2 3 0 1 ¿ e 3 3 2 3 ( ( e 3 3 ) ( e 3 x 9 ) 0 1 ) = e 3 3 2 3 ( ( e 3 3 ) 1 9 ( e 3 1 ) ) = e 3 3 2 e 3 9 + 2 27 ( e 3 1 ) = ¿ ¿ e 3 3 2 e 3 9 + 2 e 3 27 2 27 = ¿ 5 e 3 27 2 27 24. Evaluate the integral. 2 1 ln x xdx u = ln x ,du = 1 x dx ,dv = xdx ,v = x 2 2 1 2 ln x xdx = ln x ( x 2 2 ) 1 2 ( x 2 2 ) ( 1 x ) dx = ¿ x ln xdx = x 2 2 ln x 1 2 xdx ¿ x ln xdx = x 2 2 ln x x 2 4 + C 1 2 x ln xdx = ( x 2 2 ln x x 2 4 ) 1 2 = ( x 2 2 ln x x 2 4 ) 1 2 = ( 2 2 2 ln 2 2 2 4 ) ( 1 2 2 ln 1 1 2 4 ) ¿ 2ln 2 3 4 = ln 4 3 4 WA 2, p. 2
Brandon Moyer Calculus II (MAT-232-GS001) Written Assignment 2 46. Evaluate the integral using integration by parts and substitution.
Section 6.3 8. Evaluate the integral.