# Integration Question

1. Jan 24, 2005

### courtrigrad

Is $$\int^2_1 2x + x + 1 dx= (1)(3) + \frac{1}{2}(1)(3) + 1$$?

2. Jan 24, 2005

### dextercioby

Why doncha tell us how u came up with those numbers...??

Daniel.

3. Jan 24, 2005

### Ryoukomaru

$$\int^2_1 2x + x + 1 dx = \int^2_1 3x + 1 dx = \left[\frac{3}{2}x^2 + x \right]^2_1$$
When you solve that i think u get 5.5

Though i suspect that original equation would be
$$\int^2_1 2x^2 + x + 1 dx$$ or why would they give it to you in that form ... 2x + x ?

Last edited: Jan 24, 2005
4. Jan 24, 2005

### courtrigrad

They were trying to illustrate the basic integrals

$$\int^b_a x dx = \frac{1}{2}(b-a)(b+a)$$ and $$\int^b_a x^2 dx = \frac {1}{3}(b^3 - a^3)$$

$$\int^b_a c dx = c(b-a)$$

This is how I got those numbers

5. Jan 24, 2005

### Ryoukomaru

You answer is correct.

6. Jan 24, 2005

### vincentchan

your answer is absolutely correct. (if you sure the first term is 2x instead of 2x^2 as ryoukomaru stated earlier)

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