Basic Definite Integration, area under the curve

[FaraDazed]

Homework Statement

Evaluate:

A:
<br /> \int^{16}_0 (\sqrt{x} - 1)\,dx<br />

B:
<br /> \int^4_1 \frac{2}{\sqrt{x}}\,dx<br />

C:
<br /> \int^6_2 \frac{4}{x^3}\,dx<br />

Homework Equations

n/a

The Attempt at a Solution

Its part C which I think I have done wrong, but the others could be too.

Part A:
<br /> \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<br />

Part B:
<br /> \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<br />

C:
<br /> \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<br />

 

[synkk]

Homework Statement

Evaluate:

A:
<br /> \int^{16}_0 (\sqrt{x} - 1)\,dx<br />

B:
<br /> \int^4_1 \frac{2}{\sqrt{x}}\,dx<br />

C:
<br /> \int^6_2 \frac{4}{x^3}\,dx<br />

Homework Equations

n/a

The Attempt at a Solution

Its part C which I think I have done wrong, but the others could be too.

Part A:
<br /> \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<br />

Part B:
<br /> \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<br />

C:
<br /> \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<br />

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

apply the limits again

 

[FaraDazed]

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

apply the limits again

Ah yeah, don't know what i was thinking there!

Should it be...
<br /> [\frac{80}{3}]-[0] = \frac{80}{3} sq\,units<br />

And yeah 4/9 is what I meant its just the way it came out on my calc I couldn't work out what fraction produced that recurring decimal :)

Thanks for your help.