State the Fundamental Theorem of Calculus and use it to differentiate the following functions with respect to x?
F(x) =x^3 S 1 (1 − t)sin(t^2) − e^t dt
G(x) =x^2 S 0 4t^2 − sin(t) −√ t − 2 dt.
*The S is the long S shape at the start of the question.
For the first one I got 1-x^3 (Sin)(x^3)^2 - e^x^3
For the second I got 4(x^2)^2 - Sin(x^2)-Sqrt(x^2 - 2)
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For the first one, you forgot to multiply what you got by the derivative of x^3, which is 3x^2. So therefore your answer should be 3x^2 * ((1 - x^3)*sin(x^6) - e^(x^3)). You made the same mistake on the second one as well, multiply your answer by 2x.
Given: I interpret it to mean as follows. Please tell me if I'm mistaken.
F(x) = x^3*â«(0 to 1) [(1 â t)sin(t^2) â e^t] dt .... Integral from 0 to 1
G(x) = x^2*â«(0 to 1) [4t^2 â sin(t) â â Ì (t â 2)] dt
(1)
For F(x)
F(x) = x^3*â«(0 to 1) [(1 â t)sin(t^2) â e^t] dt
= x^3*[â«(0 to 1)dt â â«(0 to 1)t sin(t^2)dt â â«(0 to 1)e^t dt]
For the first integral term
â«(0 to 1) dt
= t | (0 to 1)
= 1 - 0
= 1
For the second integral term
Let u=t^2
du = 2t dt
(1/2)du = t dt
â«(0 to 1) t sin(t^2) dt
= â«(0 to 1) (1/2)sin(u) du
= (1/2)[-cos(u)] | (0 to 1)
â (1/2)[-(0.5403-1)]
â 0.2299
For the third integral term
â«(0 to 1) e^t dt
= e^t | (0 to 1)
â 1.7183
So F(x) is
F(x) â x^3*(1 - 0.2299 - 1.7183)
â -0.9482x^3
(2)
For G(x)
G(x) = x^2*â«(0 to 1) [4t^2 â sin(t) â â Ì (t â 2)] dt
= x^2*[â«(0 to 1)4t^2dt â â«(0 to 1)sin(t)dt â â«(0 to 1)â Ì (t â 2)dt]
For the first integral term
â«(0 to 1) 4t^2 dt
= (4/3)t^3 | (0 to 1)
= 4/3
â 1.3333
For the second integral term
â«(0 to 1) sin(t) dt
= -cos(t) | (0 to 1)
â 0.4597
For the third integral term
Let u=tâ2
du = dt
u = -2 to -1
â«(0 to 1) â Ì (t â 2) dt
= â«(-2 to -1) â Ì u du
= (2/3)u^(3/2) | (-2 to -1)
= (2/3)[(-1)^(3/2) - (-2)^(3/2)]
= (2/3)[-i - (-2)â Ì 2*i]
â 2.1213i
So G(x) is
G(x) â x^2(1.3333 - 0.4597 - 2.1213i)
â (0.8736 - 2.1213i)x^2