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Substitution:
u = t³
du = 3 t² dt
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∫ 3 t² [ cos ( t³ ) + 2 ] dt =
∫ [ cos ( t³ ) + 2 ] ∙ 3 t² dt =
∫ [ cos ( u ) + 2 ] du =
∫ cos ( u ) du + 2 ∫ du =
sin ( u ) + 2 u + C =
sin ( t³ ) + 2 t³ + C
F(x) =
x
∫ 3 t² [ cos ( t³ ) + 2 ] dt =
- 2
x
[ sin ( t³ ) + 2 t³ ] =
- 2
sin ( x³ ) + 2 x - [ sin ( - 2 ³ ) + 2 ∙ ( - 2 )³ ] =
sin ( x³ ) + 2 x³ - [ sin ( - 8 ) + 2 ∙ ( - 8 ) ] =
sin ( x³ ) + 2 x³ - [ - sin ( 8 ) - 16 ] =
sin ( x³ ) + 2 x³ + sin ( 8 ) + 16
F(x) = 2 x³ + sin ( x³ ) + 16 + sin ( 8 )
F'(x) = ( 2 x³ )' + [ sin ( x³ ) ] ' + ( 16 )' + [ sin ( 8 ) ]'
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Using the chain rule:
[ sin ( x³ ) ] ' = 3 x² ∙ sin ( x³ )
______________________
F'(x) = 6 x² + 3 x² ∙ sin ( x³ ) + 0 + 0 = 3 x² [ 2 + sin ( x³ ) ]
F(x) = integral (-2, x) 3t^2 (cos(t^3) +2) dt
Using u(t) = t^3 , find an equivalent equation which consists of the new definite integral.
Use your new expression to find F'(x) using Leibniz's Rule
2 answers
holy cow thank you so much !! :]