`y = int_(1 - 3x)^1((u^3)/(1 + u^2))du` Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

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gsarora17 | (Level 2) Associate Educator

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`y=int_(1-3x)^1(u^3/(1+u^2))du`

Let t=1-3x

`dt/dx=-3`

`dy/dx=dy/dt*dt/dx`

`y'=d/dxint_(1-3x)^1(u^3/(1+u^2))du`

`=d/dtint_t^1(u^3/(1+u^2))du.dt/dx`

`=-d/dtint_1^t(u^3/(1+u^2))du.dt/dx`

`=-(t^3/(1+t^2))*(-3)`

`=(3t^3)/(1+t^2)`

plug back the value of t,

`=(3(1-3x)^3)/(1+(1-3x)^2)`