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
To find an equivalent equation for the given integral using the substitution u = t^3 (u-substitution method), proceed as follows:
1. Substitute u = t^3. This implies t = u^(1/3) and dt = (1/3)u^(-2/3) du.
2. Rewrite the given bounds of integration (-2, x) in terms of u.
When x = -2, u = (-2)^3 = -8.
When x = x (which is an arbitrary value), u = x^3.
Now, plug in the new expression for t, dt, and the new bounds into the given integral:
F(x) = ∫(-2, x) 3t^2(cos(t^3) + 2) dt
= ∫(-8, x^3) 3(u^(2/3))(cos(u) + 2) * (1/3)u^(-2/3) du
= ∫(-8, x^3) (u^(2/3))(cos(u) + 2) * u^(-2/3) du
= ∫(-8, x^3) cos(u) + 2 du
= ∫(-8, x^3) cos(u) du + ∫(-8, x^3) 2 du
= sin(u)∣(-8, x^3) + 2u∣(-8, x^3)
= sin(x^3) - sin(-8) + 2(x^3 - (-8))
= sin(x^3) + 2x^3 + 14
Now, to find F'(x) using Leibniz's Rule, differentiate F(x) with respect to x:
F'(x) = (d/dx) (sin(x^3) + 2x^3 + 14)
= 3x^2 cos(x^3) + 6x^2
Therefore, F'(x) = 3x^2 cos(x^3) + 6x^2.
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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 )
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F'(x) = 6 x² + 3 x² ∙ sin ( x³ ) + 0 + 0 = 3 x² [ 2 + sin ( x³ ) ]