# Calculating the integral of x^n e^(2x) from 0 to 1

1. Aug 25, 2006

### ultima9999

Hey, I got a couple of problems on integration that I can't seem to figure out.

1. Suppose $$I_n = \int_{0}^{1} x^n e^{2x} dx$$ Evaluate $$I_n$$ in terms of $$I_{n-1}$$ for any natural number n

2. Suppose $$I_n = \int_{1}^{e} \left[\ln x\right]^n dx$$ Evaluate $$I_n$$ in terms of $$I_{n-1}$$ for any natural number n

Not sure what to do with these. Do I need to integrate $$I_{n-1}$$? What do I put into $$I_n$$ after I integrate $$I_{n-1}$$??

2. Aug 25, 2006

### TD

With

$$I_n = \int_{0}^{1} x^n e^{2x} dx$$

then

$$I_{n-1} = \int_{0}^{1} x^{n-1} e^{2x} dx$$

Use integration by parts to get a relation between I_n and I_n-1.

3. Aug 25, 2006

### ultima9999

Yup, I did that and I got $$I_n = \frac{1}{2}x^ne^{2x} - \frac{n}{2}I_{n-1}$$

For the second question, I let u = ln x, but what would I let dv/dx equal to?

4. Aug 25, 2006

### TD

It's a bit unclear to me what you mean with dv/dx, but do you mean identifying the factors for integration by parts?

Remember that one can integrate ln(x)dx by looking at it as ln(x).1dx and taking u = ln(x) and v = 1, dv = dx.

5. Aug 25, 2006

### ultima9999

Yup! $$I_n = x[\ln x]^n - nI_{n-1}$$