Use 2nd Fundamental Theorem of Calculus to find derivative of f(x) = integral of 2x^2 to x-5 of square root of Sin(t)dt.

Please help.

I'm sorry, but the integrand

"2x^2 to x-5 of square root of Sin(t)dt."

does not make sense.

In general, the derivative of an indefinite integral of a function is the function itself.

Ok, so if I change it to 2x^2 to x-5 of square root of Sin(x)dx, does it make sense?

We haven't done any examples with an x in both the top and bottom of the integral sign.
Thanks

To find the derivative of the function f(x) = ∫(x-5) to x 2√sin(t) dt using the Second Fundamental Theorem of Calculus, follow these steps:

Step 1: Rewrite the function f(x) using the First Fundamental Theorem of Calculus.

The First Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on the interval [a, b], then ∫(a) to (b) f(x) dx = F(b) - F(a).

In our case, let's find an antiderivative of the integrand 2√sin(t). Notice that the integrand contains a composite function, √sin(t). To integrate this function, we can use the chain rule.

Let's denote F(t) as an antiderivative of 2√sin(t). Applying the chain rule, we have F(t) = ∫ 2√sin(t) dt = ∫ 2(sin(t))^1/2 dt. Let u = sin(t), then du = cos(t) dt. Substituting these values, we get:

F(t) = ∫ 2u^1/2 (1/cos(t)) du
= ∫ 2u^1/2 (1/cos(t)) (du/dt) dt
= ∫ 2u^1/2 (1/cos(t)) cos(t) dt
= ∫ 2u^1/2 dt
= 2 ∫ u^1/2 dt.

Taking the integral, we get:

F(t) = 2 * (2/3) * u^3/2 + C
= (4/3) * u^3/2 + C
= (4/3) * (sin(t))^3/2 + C.

Therefore, an antiderivative, or primitive, of the integrand 2√sin(t) is F(t) = (4/3) * (sin(t))^3/2 + C, where C is the constant of integration.

Step 2: Apply the Second Fundamental Theorem of Calculus to find the derivative of f(x).

The Second Fundamental Theorem of Calculus states that if f(x) is a continuous function on the interval [a, b] and F(x) is an antiderivative of f(x), then d/dx ∫(a) to (x) f(t) dt = f(x).

In our case, f(x) = ∫(x-5) to x 2√sin(t) dt. By substituting the antiderivative F(t) = (4/3) * (sin(t))^3/2 + C into this equation, we get:

f(x) = F(x) - F(x-5)
= (4/3) * (sin(x))^3/2 + C - (4/3) * (sin(x-5))^3/2 + C
= (4/3) * (sin(x))^3/2 - (4/3) * (sin(x-5))^3/2 + 2C.

Now, to find the derivative of f(x), simply differentiate both sides of the equation with respect to x:

d/dx [f(x)] = d/dx [(4/3) * (sin(x))^3/2 - (4/3) * (sin(x-5))^3/2 + 2C].

Differentiating the first term with respect to x:

d/dx [(4/3) * (sin(x))^3/2] = (2/3) * 3(sin(x))^1/2 * cos(x)
= 2(sin(x))^1/2 * cos(x).

Differentiating the second term with respect to x:

d/dx [(4/3) * (sin(x-5))^3/2] = (2/3) * 3(sin(x-5))^1/2 * cos(x-5)
= 2(sin(x-5))^1/2 * cos(x-5).

Since C is a constant, the derivative of 2C with respect to x is 0.

Therefore, the derivative of f(x) is:

d/dx [f(x)] = 2(sin(x))^1/2 * cos(x) - 2(sin(x-5))^1/2 * cos(x-5).

This is the derivative of the function f(x) using the Second Fundamental Theorem of Calculus.