How do you find the derivative of #G(x)=int (t^2-3t)dt# from #[1,x^2]#?

Question

Evelyn Agard · Accepted Answer

# G'(x) = 2x^5-6x^3 #

We have: # G(x) = int_1^(x^2) (t^2-3t) \ dt # If we let # F(x) = int_a^x (t^2-3t) \ dt # Then from the first part of the Fundamental Theorem of Calculus, we have: # F'(x) = x^2-3x " "....... [1] # And from the second part of the Fundamental Theorem of Calculus, we have: # int_a^x (t^2-3t) \ dt = F(x) - F(a) " "....... [2] # Therefore, # G(x) = F(x^2) - F(a) " "# from [2] Differentiating wrt #x# we get # \ \d/dx G(x) = d/dx F(x^2) - d/dx F(a) # # :. G'(x) = F'(x^2)d/dx (x^2) - 0 " "# chain rule # " "= ((x^2)^2-3(x^2)) (2x) " "# from [1] # " "= (x^4-3x^2) (2x) # # " "= 2x^5-6x^3 #

Isabella Ackerman · Answer

undefined To find the derivative of $ G(x) = \int_{1}^{x^2} (t^2 - 3t) dt $, we can use the Fundamental Theorem of Calculus and the chain rule.

First, we find the antiderivative of $ t^2 - 3t $, which is $ \frac{1}{3}t^3 - \frac{3}{2}t^2 $.

Next, we evaluate this antiderivative at the upper and lower limits of integration and subtract.

So, $ G(x) = \left[ \frac{1}{3}t^3 - \frac{3}{2}t^2 \right]_{1}^{x^2} $

Evaluating at the upper limit, we get $ \frac{1}{3}x^6 - \frac{3}{2}x^4 $.

Evaluating at the lower limit, we get $ \frac{1}{3} - \frac{3}{2} $.

Subtracting these results, we get $ G(x) = \frac{1}{3}x^6 - \frac{3}{2}x^4 - \frac{1}{3} + \frac{3}{2} $.

Now, we differentiate $ G(x) $ with respect to $ x $ to find the derivative:

$ G'(x) = \frac{d}{dx} \left( \frac{1}{3}x^6 - \frac{3}{2}x^4 - \frac{1}{3} + \frac{3}{2} \right) $

$ G'(x) = 2x^5 - 6x^3 $.

So, the derivative of $ G(x) $ is $ 2x^5 - 6x^3 $.