234k views
1 vote
Given the existence of an antiderivative F(x)=∫0xe2tdt, find the value of x such that F(x)=3. Justify your work.

User Cvacca
by
7.5k points

1 Answer

6 votes

Final answer:

To find the value of x where F(x)=3 for the antiderivative F(x) = ∫_0^x e^{2t}dt, we set up and solve the equation (e^{2x} - 1)/2 = 3, resulting in x being equal to ln(7)/2.

Step-by-step explanation:

The student is asked to find the value of x such that the antiderivative F(x) = ∫_0^x e^{2t}dt equals 3. To solve for x, we differentiate F(x) by applying the Fundamental Theorem of Calculus, which tells us that F'(x) = ​e^{2x}. We then integrate e^{2x} from 0 to x and set the equation equal to 3:

​F(x) = ∫_0^x e^{2t}dt = 3

Since ∫_0^x e^{2t}dt = (e^{2x} - 1)/2, we have:

(e^{2x} - 1)/2 = 3

Solving this equation for x, we get:

  1. e^{2x} - 1 = 6
  2. e^{2x} = 7
  3. 2x = ​ln(7)
  4. x = ​ln(7)/2

Thus, the value of x that makes F(x) equal to 3 is ln(7)/2.

User M Omayr
by
8.6k points