Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
g(x) = e^(3t^2) − 3t dt from 4 to x
Hmmm. Your question is ill worded. I suspect you mean g(x) is the integral of e^(3t^2)-3t dt from 4 to x.
Otherwise, we'd need the integral of e^(3t^2) which is not elementary.
So, with that in mind,
dg/dx = e^(3x^2)-3x
To find the derivative of the function g(x) = e^(3t^2) − 3t, we can use Part 1 of the Fundamental Theorem of Calculus.
The theorem states that if a function f(x) is continuous on an interval [a, b] and F(x) is any antiderivative of f(x) on [a, b], then:
∫[a to b] f(x) dx = F(b) - F(a)
In this case, we want to find the derivative of the function g(x) = e^(3t^2) − 3t, where the variable of integration is t. However, we are given the limits of integration as 4 to x.
Therefore, to use the Fundamental Theorem of Calculus, we need to rewrite the expression to have the same limits of integration as given.
We can rewrite the integral as:
∫[4 to x] (e^(3t^2) − 3t) dt
Now, we can use the Fundamental Theorem of Calculus Part 1 to find the derivative.
The derivative of g(x), denoted as g'(x), is given by:
g'(x) = d/dx [∫[4 to x] (e^(3t^2) − 3t) dt]
Applying the Fundamental Theorem of Calculus, we know that the derivative of the integral is just the integrand evaluated at the upper limit of integration. Therefore:
g'(x) = (e^(3x^2) − 3x)
To find the derivative of the function using Part 1 of the Fundamental Theorem of Calculus, we need to find the antiderivative of the function first.
The antiderivative of g(x) is denoted as G(x). Since the function is a definite integral from the constant 4 to x, we evaluate the antiderivative at the upper limit x and subtract the value of the antiderivative at the lower limit 4.
Let's find the antiderivative first.
The function g(x) = e^(3t^2) - 3t.
To find the antiderivative of e^(3t^2), we need to use the chain rule. The derivative of e^(3t^2) is 2t * 3e^(3t^2), where 2t comes from the chain rule.
So, the antiderivative of e^(3t^2) is (1/3) * 2te^(3t^2).
To find the antiderivative of -3t, we use the power rule, which states that ∫x^n dx = (1/(n+1)) * x^(n+1).
So, the antiderivative of -3t is (-3/2) * t^2.
Now, we can put these antiderivatives together to find G(x).
G(x) = (1/3) * 2te^(3t^2) - (3/2) * t^2.
Now that we have G(x), we can apply Part 1 of the Fundamental Theorem of Calculus, which states that if F(x) is the antiderivative of f(x), then ∫[a,b] f(x) dx = F(b) - F(a).
In this case, the derivative of the function g(x) is the derivative of G(x).
So, G'(x) = ((1/3) * 2te^(3t^2))' - ((3/2) * t^2)'
Differentiating the terms of G(x) with respect to x:
G'(x) = (2/3) * e^(3t^2) + 6t^2 - 3t
Therefore, the derivative of the function g(x) is G'(x) = (2/3) * e^(3t^2) + 6t^2 - 3t.