Use the graph of f(t) = 2t + 3 on the interval [-3, 6] to write the function F(x), where F(x)= ∫f(t) dt where a=3 b=x.
F(x) = 2x^2 + 6x
F(x) = 2x + 3
F(x) = x^2 + 3x + 54
F(x) = x^2 + 3x - 18
Honestly have no idea where to start. Do i take the derivative of that or what?
surely you can integrate
∫2t+3 dt
The tricky part is what to do with the limits. That means you want
F(x) = ∫[3,x] 2t+3 dt
Now just evaluate normally, but using x as the upper limit, instead of a number. Then you have
F(x) = t^2+3t [3,x]
= (x^2+3x)-(3^2+3*3)
= x^2+3x-18
Well, don't worry! It's time to bring out my clown calculator and solve this problem with a smile. To find the function F(x), you'll have to integrate the given function f(t) over the interval [-3, x].
First, let's solve the integral of f(t) = 2t + 3.
Integrating 2t with respect to t gives us t^2, and integrating 3 with respect to t gives us 3t. So, the integral of f(t) is:
∫f(t) dt = ∫(2t + 3) dt = t^2 + 3t
Now, to find F(x), we need to evaluate this antiderivative between the limits a = 3 and b = x:
F(x) = ∫[from 3 to x] (2t + 3) dt
To evaluate this, we need to substitute the limits of integration into the antiderivative.
Let's start by evaluating the antiderivative at the upper limit, x:
F(x) = (x^2 + 3x) - (3^2 + 3 * 3)
Simplifying that, we get:
F(x) = x^2 + 3x - 18
Congratulations, you've got it! No need to take derivatives, just integrate and evaluate between the limits. I hope this helps, and remember, math is always funnier with a clown around!
To find the function F(x), which represents the integral of f(t) with respect to t from 3 to x, you need to find the antiderivative (or indefinite integral) of f(t) and then evaluate it at the given limits.
The function f(t) = 2t + 3 represents a straight line with a slope of 2 and y-intercept of 3.
To find the antiderivative of f(t), you need to integrate each term separately.
∫(2t + 3) dt = ∫2t dt + ∫3 dt
The antiderivative of 2t with respect to t is t^2 (divided by 2), and the antiderivative of 3 with respect to t is 3t:
∫(2t) dt = t^2
∫3 dt = 3t
Now, add these antiderivatives together:
F(x) = (t^2/2) + 3t
To find F(x) on the interval [3, x], we need to evaluate F(x) at the upper limit (x) and subtract the value of F(x) at the lower limit (3).
Evaluate F(x) at x:
F(x) = (x^2/2) + 3x
Evaluate F(x) at 3:
F(3) = (3^2/2) + 3(3) = 9/2 + 9 = 9/2 + 18/2 = 27/2
Finally, subtract the value of F(x) at 3 from the value of F(x) at x to get the function F(x):
F(x) = (x^2/2) + 3x - 27/2
To find the function F(x), which represents the integral of f(t) from 3 to x, you need to use the Fundamental Theorem of Calculus. Here are the steps to find F(x):
1. Find the antiderivative of f(t):
The antiderivative of f(t) = 2t + 3 is F(t) = t^2 + 3t + C, where C is the constant of integration.
2. Evaluate the antiderivative at the upper limit x:
Plug in x into the antiderivative F(t) = t^2 + 3t + C to get F(x) = x^2 + 3x + C.
3. Evaluate the antiderivative at the lower limit 3:
Plug in 3 into the antiderivative F(t) = t^2 + 3t + C to get F(3) = 3^2 + 3 × 3 + C = 18 + 9 + C = 27 + C.
4. Subtract the two results to eliminate the constant of integration:
F(x) - F(3) = (x^2 + 3x + C) - (27 + C) = x^2 + 3x + C - 27 - C = x^2 + 3x - 27.
Therefore, the function F(x) = x^2 + 3x - 27 represents the integral of f(t) = 2t + 3 from 3 to x.