How do you use the second fundamental theorem of Calculus to find the derivative of given #int sqrt(6 + r^3) dr# from #[7, x^2]#?

Question

How do you use the second fundamental theorem of Calculus to find the derivative of given #int  sqrt(6 + r^3) dr# from #[7, x^2]#?

Charles Anctil · Accepted Answer

#d/dx int_7^(x^2)sqrt(6+r^3)dr=2xsqrt(6+x^6)#

Let #F(r) + C = intsqrt(6+r^3)dr#, that is, the function such that #d/(dr)F(r) = sqrt(6+r^3)#. The second fundamental theorem of calculus states that

#int_a^bsqrt(6+r^3)dr = F(b)-F(a)#

With that, we have

#d/dx int_7^(x^2)sqrt(6+r^3)dr = d/dx(F(x^2)-F(7))#

#=d/dxF(x^2) - d/dxF(7)#

#=sqrt(6+(x^2)^3)(d/dxx^2)-0#

#=2xsqrt(6+x^6)#

where in the second to last step we used the chain rule along with the fact that #F(7)# is a constant, and thus has a derivative of #0#.

Ezra Adams · Answer

undefined To use the second fundamental theorem of calculus to find the derivative of the given integral $ \int_{7}^{x^2} \sqrt{6 + r^3} \, dr $, follow these steps:

1. Find the antiderivative of the integrand with respect to $ r $.
2. Substitute the upper limit $ x^2 $ into the antiderivative.
3. Substitute the lower limit $ 7 $ into the antiderivative.
4. Subtract the result of substituting the lower limit from the result of substituting the upper limit to find the derivative.

Here's the solution:

1. The antiderivative of $ \sqrt{6 + r^3} $ with respect to $ r $ is $ \frac{2}{9}(6 + r^3)^{3/2} + C $, where $ C $ is the constant of integration.

2. Substitute the upper limit $ x^2 $ into the antiderivative:
$$ \frac{2}{9}(6 + (x^2)^3)^{3/2} + C = \frac{2}{9}(6 + x^6)^{3/2} + C $$

3. Substitute the lower limit $ 7 $ into the antiderivative:
$$ \frac{2}{9}(6 + 7^3)^{3/2} + C = \frac{2}{9}(6 + 343)^{3/2} + C = \frac{2}{9}(349)^{3/2} + C $$

4. Subtract the result of substituting the lower limit from the result of substituting the upper limit:
$$ \frac{2}{9}(6 + x^6)^{3/2} + C - \frac{2}{9}(349)^{3/2} - C = \frac{2}{9}(6 + x^6)^{3/2} - \frac{2}{9}(349)^{3/2} $$

So, the derivative of $ \int_{7}^{x^2} \sqrt{6 + r^3} \, dr $ is $ \frac{2}{9}(6 + x^6)^{3/2} - \frac{2}{9}(349)^{3/2} $.