What is #f(x) = int e^(2x-1)-e^(1-2x)+e^x dx# if #f(2) = 3 #?
The function
By signing up, you agree to our Terms of Service and Privacy Policy
To find (f(x) = \int (e^{2x-1} - e^{1-2x} + e^x) , dx) given that (f(2) = 3), we integrate the function term by term and then apply the initial condition to solve for the constant of integration.
- Integrate (e^{2x-1} - e^{1-2x} + e^x):
[ \int e^{2x-1} , dx - \int e^{1-2x} , dx + \int e^x , dx ]
For (\int e^{2x-1} , dx), use a substitution (u = 2x - 1 \Rightarrow du = 2dx):
[ \frac{1}{2}\int e^u , du = \frac{1}{2}e^u + C_1 = \frac{1}{2}e^{2x-1} ]
For (\int e^{1-2x} , dx), use a substitution (v = 1-2x \Rightarrow dv = -2dx):
[ -\frac{1}{2}\int e^v , dv = -\frac{1}{2}e^v + C_2 = -\frac{1}{2}e^{1-2x} ]
For (\int e^x , dx):
[ \int e^x , dx = e^x + C_3 ]
Combining these:
[ f(x) = \frac{1}{2}e^{2x-1} - \frac{1}{2}e^{1-2x} + e^x + C ]
- Apply the initial condition (f(2) = 3):
[ f(2) = \frac{1}{2}e^{2(2)-1} - \frac{1}{2}e^{1-2(2)} + e^{2} + C = 3 ]
[ = \frac{1}{2}e^{3} - \frac{1}{2}e^{-3} + e^{2} + C = 3 ]
Calculate the values:
[ C = 3 - \left(\frac{1}{2}e^{3} - \frac{1}{2}e^{-3} + e^{2}\right) ]
Thus, the function (f(x)) is:
[ f(x) = \frac{1}{2}e^{2x-1} - \frac{1}{2}e^{1-2x} + e^x + 3 - \left(\frac{1}{2}e^{3} - \frac{1}{2}e^{-3} + e^{2}\right) ]
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- How do you integrate #int 1/sqrt(x^2-16x+37) # using trigonometric substitution?
- How do you integrate #int4sec4x*tan4x*sec^4 4x# using substitution?
- What is #f(x) = int xe^x-xsqrt(x^2+2)dx# if #f(0)=-2 #?
- How do you integrate #int x/(x-6) dx# using partial fractions?
- How do you integrate #int x^3 e^(x^2 ) dx # using integration by parts?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7