Integration of √e^x and 1/e^(x-1)

Integration is a fundamental operation in calculus that involves finding the antiderivative of a given function. In this case, we are asked to find the integral of √e^x and 1/e^(x-1). Let's break down the process step by step.

1. Integral of √e^x:
To find the integral of √e^x, we can use the substitution method. Let's substitute u = e^x, which means du = e^x dx.

∫√e^x dx = ∫√u du
Now, we have transformed the integral into a simpler form.

2. Integrating ∫√u du:
To integrate ∫√u du, we can use the power rule for integration. When we have an expression of the form ∫u^n du, where n is not equal to -1, the integral is given by:

∫u^n du = (u^(n+1))/(n+1) + C
where C is the constant of integration.

Applying the power rule to our integral, we have:

∫√u du = (u^(1/2))/(1/2) + C
Simplifying this expression, we get:

∫√u du = 2√u + C

3. Substituting back:
Now, we can substitute back u = e^x into our expression for the integral:

∫√e^x dx = 2√e^x + C

4. Integral of 1/e^(x-1):
To find the integral of 1/e^(x-1), we can use the power rule again. We know that:

∫u^n du = (u^(n+1))/(n+1) + C

Applying this rule to our integral, we have:

∫1/e^(x-1) dx = ∫e^(-(x-1)) dx

Using the substitution method again, let's substitute v = -(x-1), which means dv = -dx.

∫e^(-(x-1)) dx = ∫e^v (-dv)
Now, we have transformed the integral into a simpler form.

∫e^v (-dv) = -∫e^v dv
Since the integral of e^v is simply e^v, we have:

-∫e^v dv = -e^v + C

5. Substituting back:
Finally, we can substitute back v = -(x-1) into our expression for the integral:

∫1/e^(x-1) dx = -e^(-(x-1)) + C

In conclusion, the integral of √e^x is 2√e^x + C, and the integral of 1/e^(x-1) is -e^(-(x-1)) + C.

Can you write it integrand and attach the image also?