How do you use the second fundamental theorem of Calculus to find the derivative of given #int sin^3(2t-1) dt# from #[x-1, x]#?

Question

Evelyn Agard · Accepted Answer

Hence we need to find the derivative of

#int_(x-1)^[x] sin^3(2t-1)dt#

So we use the Leibniz Integral Rule which in our case becomes

#d/dx[int_(a(x))^(b(x)) f(t)dt]=f(b(x))*(db(x))/dx-f(a(x))*(d(a(x)))/dx#

Hence the derivative for #b(x)=x# ,#a(x)=x-1# ,#f(t)=sin^3(2t-1)#

is

#d/dx( int_(x-1)^x sin^3(2 t-1) dt) = sin^3(2x-1)*(x)'-sin^3(2(x-1)-1)*(x-1)'=sin^3(2x-1)-sin^3(2x-3)#

William Abdalla · Answer

undefined To use the second fundamental theorem of calculus to find the derivative of the integral $ \int_{x-1}^{x} \sin^3(2t-1) \, dt $, we first need to express the integral as a function of $ x $.

Let $ F(x) = \int_{x-1}^{x} \sin^3(2t-1) \, dt $.

Then, by the second fundamental theorem of calculus, the derivative of $ F(x) $ is given by:

$$ F'(x) = \sin^3(2x-1) \cdot \frac{d}{dx}(2x-1) - \sin^3(2(x-1)-1) \cdot \frac{d}{dx}(2(x-1)-1) $$

Simplify this expression to find the derivative.