How do you use the Fundamental Theorem of Calculus to find the derivative of #int xsin(πx) # from 1 to x^2?

Question

Cameron Alexander · Accepted Answer

#d/dxI = 2x^3*sin(pi*x^2)#

Strictly speaking you don't, because you can't have the variable you're integrating show up in the limits of integration. Assuming you meant something like

#I = int_1^(x^2)tsin(pit)dt#

We see that the fundamental theorem of calculus states that

# int_a^bf(x)dx = F(b) - F(a)#

Where #f(x) = d/dxF(x)# so using the Fundamental Theorem of Calculus we have

#I = F(x^2) - F(1)#

Now, we just differentiate both sides

#d/dxI = d/dxF(x^2) - d/dxF(1)#

#F(1)# is a constant so its derivative is #0#

#d/dxI = d/dxF(x^2)#

Using the chain rule:

#u = x^2# and #(du)/dx = 2x#

#d/dxI = d/(du)F(u)*(du)/dx = 2x*d/(du)F(u)#

However we know that #F(u)# is a function such that its derivative is #u*sin(pi*u)#

#d/dxI = 2x*u*sin(pi*u)#

But since #u = x^2#

#d/dxI = 2x*x^2*sin(pi*x^2) = 2x^3*sin(pi*x^2)#

Samantha Adkison · Answer

undefined To use the Fundamental Theorem of Calculus to find the derivative of $ \int_{1}^{x^2} x \sin(\pi x) \, dx $, follow these steps:

1. Evaluate the antiderivative of the integrand $ x \sin(\pi x) $:
   $$ \int x \sin(\pi x) \, dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi} + C $$

2. Substitute the upper limit ($ x^2 $) into the antiderivative expression:
   $$ -\frac{x^2 \cos(\pi x^2)}{\pi} + \frac{\sin(\pi x^2)}{\pi} + C $$

3. Substitute the lower limit ($ 1 $) into the antiderivative expression:
   $$ -\frac{1 \cdot \cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi} + C $$

4. Simplify:
   $$ -\frac{\cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi} + C $$

5. Take the derivative of the expression obtained in steps 2 to 4 with respect to $ x $:
   $$ \frac{d}{dx} \left( -\frac{x^2 \cos(\pi x^2)}{\pi} + \frac{\sin(\pi x^2)}{\pi} + C \right) $$

6. Simplify the derivative:
   $$ -\frac{2x \cos(\pi x^2)}{\pi} - 2x \sin(\pi x^2) $$

Therefore, the derivative of $ \int_{1}^{x^2} x \sin(\pi x) \, dx $ is $ -\frac{2x \cos(\pi x^2)}{\pi} - 2x \sin(\pi x^2) $.