How do you use part 1 of the Fundamental Theorem of Calculus to find the derivative of #y=int(sqrtt sintdt)# from #[x^3, sqrtx]#?

Question

Samantha Adkison · Accepted Answer

You need one constant limit of integration to apply FTC Part 1.

So rewrite the integral:

#y=int_(x^3)^sqrtx (sqrtt sintdt) = int_(x^3)^1 (sqrtt sintdt) +int_1^sqrtx (sqrtt sintdt) # ,

So # y = - int_1^(x^3) (sqrtt sintdt) +int_1^sqrtx (sqrtt sintdt)#

Now use FTC 1 and the Chain Rule to get:

#-sqrt(x^3)sin(x^3)3x^2 + sqrtsqrtx sin (sqrtx) 1/2sqrtx#

#= -3x^3sqrtxsin(x^3) + root(4)x / (2sqrtx) sin(sqrtx) #

Or whatever rewrite you favor as "simplified".

Luke Alvarez · Answer

undefined To find the derivative of $ y = \int_{x^3}^{\sqrt{x}} \sqrt{t} \sin(t) \, dt $ with respect to $ x $, you can apply the Fundamental Theorem of Calculus, Part 1, which states that if $ f $ is continuous on $[a, b]$ and $ F $ is an antiderivative of $ f $ on $[a, b]$, then

$$ \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(x) $$

For this problem, first, you need to find an antiderivative of $ \sqrt{t} \sin(t) $. Then apply the Fundamental Theorem of Calculus to evaluate the derivative of the integral.