Use Fundemental Theorem of Calculus part 1 to find the derivatives of the following:
(i) F(x)=Integral ( top3, bottom x) e^-t^2 dt
(ii) G(theta)=Integral ( top cos theta, bottom sin theta) 1/1-x^2 dx
To find the derivatives of the given functions using the Fundamental Theorem of Calculus (Part 1), we need to follow a few steps.
(i) F(x) = ∫(x to 3) e^(-t^2) dt
Step 1: Rewrite the function as a definite integral.
F(x) = ∫(3) e^(-t^2) dt - ∫(x) e^(-t^2) dt
Step 2: Apply the Fundamental Theorem of Calculus (Part 1), which states that if F(x) is the definite integral of a function f(t) from a to x, then the derivative of F(x) with respect to x is f(x).
F'(x) = d/dx [∫(3) e^(-t^2) dt - ∫(x) e^(-t^2) dt]
Step 3: Evaluate the derivative of each integral using the Chain Rule and the second part of the Fundamental Theorem of Calculus.
For the first term:
d/dx [∫(3) e^(-t^2) dt] = 0 (since it does not depend on x)
For the second term:
d/dx [∫(x) e^(-t^2) dt] = e^(-x^2) * d/dx[x] = e^(-x^2)
Therefore, F'(x) = e^(-x^2)
(ii) G(θ) = ∫(cosθ to sinθ) 1/(1-x^2) dx
Step 1: Rewrite the function as a definite integral.
G(θ) = ∫(sinθ) 1/(1-x^2) dx - ∫(cosθ) 1/(1-x^2) dx
Step 2: Apply the Fundamental Theorem of Calculus (Part 1).
G'(θ) = d/dθ [∫(sinθ) 1/(1-x^2) dx - ∫(cosθ) 1/(1-x^2) dx]
Step 3: Evaluate the derivative of each integral using the Chain Rule and the second part of the Fundamental Theorem of Calculus.
For the first term:
d/dθ [∫(sinθ) 1/(1-x^2) dx] = -cosθ/(1-sin^2(θ)) * d/dθ[sinθ] = -cosθ/(1-sin^2(θ)) * cosθ
For the second term:
d/dθ [∫(cosθ) 1/(1-x^2) dx] = sinθ/(1-cos^2(θ)) * d/dθ[cosθ] = sinθ / (1-cos^2(θ)) * (-sinθ)
Therefore, G'(θ) = -cosθ/(1-sin^2(θ)) * cosθ - sinθ / (1-cos^2(θ)) * (-sinθ) = -cosθ/(1-sin^2(θ)) - sinθ / (1-cos^2(θ))