# Basic Definite Integration, area under the curve

1. May 2, 2013

1. The problem statement, all variables and given/known data
Evaluate:

A:
$$\int^{16}_0 (\sqrt{x} - 1)\,dx$$

B:
$$\int^4_1 \frac{2}{\sqrt{x}}\,dx$$

C:
$$\int^6_2 \frac{4}{x^3}\,dx$$

2. Relevant equations

n/a

3. The attempt at a solution
Its part C which I think I have done wrong, but the others could be too.

Part A:
$$\int^{16}_0 (\sqrt{x} - 1)\,dx = \int^{16}_0 (x^{\frac{1}{2}}-1)\,dx = [\frac{2}{3}x^{1.5}-1x]^{16}_0 = [112]-[0] = 112 \,sq\, units$$

Part B:
$$\int^4_1 \frac{2}{\sqrt{x}}\,dx = \int^4_1 2x^{-\frac{1}{2}}\,dx = [4x^{\frac{1}{2}}]^4_1 = [8] - [4] = 4 \,sq\, units$$

C:
$$\int^6_2 \frac{4}{x^3}\,dx = \int^6_2 4x^{-3}\,dx = [-2x^{-2}]^6_2 = [-\frac{1}{18}] - [-\frac{1}{2}] = 0.44 \,sq\, units$$

Last edited: May 2, 2013
2. May 2, 2013

### synkk

first is incorrect - others are fine (assuming you mean 4/9 for the third).

apply the limits again

3. May 2, 2013

Ah yeah, dont know what i was thinking there!

Should it be...
$$[\frac{80}{3}]-[0] = \frac{80}{3} sq\,units$$

And yeah 4/9 is what I meant its just the way it came out on my calc I couldnt work out what fraction produced that recurring decimal :)