Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function h(x)=integral [2,1/x]sin^4t dt
To find the derivative of the function h(x) = ∫[2,1/x] sin^4(t) dt using the Fundamental Theorem of Calculus, you'll need to follow a two-step process.
Step 1: Rewrite the integral in terms of x.
To do this, start by substituting the limits of integration. Since the upper limit is 1/x, you'll need to change the variable of integration from t to u, such that:
u = 1/x
Now, rewrite the integral using the variable u:
∫[2,1/x] sin^4(t) dt = ∫[2,u] sin^4(t) dt
Step 2: Apply the Fundamental Theorem of Calculus.
According to the Part 1 of the Fundamental Theorem of Calculus, if a function f(x) is defined as the integral of another function F(x), then the derivative of f(x) with respect to x is equal to F(x):
h'(x) = d/dx ∫[2,u] sin^4(t) dt
Now, differentiate F(x) with respect to x. In this case, F(x) is ∫[2,u] sin^4(t) dt:
F'(x) = d/dx ∫[2,u] sin^4(t) dt
To differentiate ∫[2,u] sin^4(t) dt, you'll need to use the chain rule of differentiation. The chain rule states that if g(x) is a composite function, such as g(x) = f(h(x)), then its derivative is given by g'(x) = f'(h(x)) * h'(x).
In this case, h(x) is the upper limit of integration, so we'll use the chain rule. Let F(u) be the antiderivative of sin^4(t) with respect to t:
F(u) = ∫ sin^4(t) dt
Now differentiate F(u) with respect to u:
F'(u) = d/du ∫ sin^4(t) dt
Finally, apply the chain rule by multiplying F'(u) by du/dx, which is the derivative of u with respect to x:
h'(x) = F'(u) * du/dx
Substituting F'(u), which is d/du of ∫ sin^4(t) dt, we get:
h'(x) = d/du ∫ sin^4(t) dt * du/dx
Now, differentiate ∫ sin^4(t) dt with respect to u:
h'(x) = ∫ d/dt (sin^4(t)) * du/dx
Differentiating sin^4(t) with respect to t gives:
h'(x) = ∫ 4sin^3(t)cos(t) * du/dx
Finally, substitute back the original variable u:
h'(x) = ∫ 4sin^3(t)cos(t) * d/dx (1/x)
Differentiating 1/x with respect to x gives:
h'(x) = ∫ 4sin^3(t)cos(t) * (-1/x^2)
Thus, the derivative of the function h(x) = ∫[2,1/x] sin^4(t) dt is h'(x) = ∫ 4sin^3(t)cos(t) * (-1/x^2).
To find the derivative of the function h(x) using the Fundamental Theorem of Calculus, we can follow these steps:
Step 1: Determine the antiderivative of the integrand:
∫sin^4(t) dt
The integral of sin^4(t) can be found by using the identity sin^2(t) = (1 - cos(2t))/2. We can rewrite sin^4(t) as (sin^2(t))^2 which is equal to ((1 - cos(2t))/2)^2.
∫sin^4(t) dt = ∫((1 - cos(2t))/2)^2 dt
Now we can expand the expression and simplify it:
= 1/4 ∫ (1 - 2cos(2t) + cos^2(2t)) dt
= 1/4 ∫ (1 - 2cos(2t) + (1 + cos(4t))/2) dt
= 1/4 ∫(3/2 - 2cos(2t) + cos(4t)/2) dt
= 3/8t - 1/4sin(2t) + 1/8sin(4t) + C
Step 2: Apply the Fundamental Theorem of Calculus by differentiating h(x) with respect to x:
h(x) = ∫[2,1/x] sin^4(t) dt
Now we can rewrite h(x) in terms of x:
h(x) = ∫[2,1/x] sin^4(t) dt
= 3/8t - 1/4sin(2t) + 1/8sin(4t) + C
where t = 1/x
Step 3: Differentiate h(x) with respect to x:
h'(x) = (d/dx) (3/8t - 1/4sin(2t) + 1/8sin(4t) + C)
Using the Chain Rule, we can express this in terms of x:
= (d/dx)(3/8(1/x) - 1/4sin(2(1/x)) + 1/8sin(4(1/x))) + C
Simplifying further, we get:
= -3/8x^2 + 1/4(sinx/x^2) - 1/2cos(2/x)(1/x^2) + 1/2cos(4/x)(1/x^2)
= -3/8x^2 + (sinx)/(4x^2) - (cos(2/x))/(2x^3) + (cos(4/x))/(2x^3)
Therefore, the derivative of the original function h(x) = ∫[2,1/x]sin^4(t) dt is given by:
h'(x) = -3/8x^2 + (sinx)/(4x^2) - (cos(2/x))/(2x^3) + (cos(4/x))/(2x^3)