# Thread: Integration questions

1. ## Integration questions

Hi
Can someone look where i have made a mistake because it doesn't match the answers in the book.

1) $\int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})$

$[\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]$

$[\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]$

$\frac{-6a^\frac{3}{2}}{3}$

the book's answer is $\frac{-5a^\frac{3}{2}}{3}$

2) $\int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}$

$6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}$

final answer is: $\frac{-38}{3}$
but book answer is $\frac{-55}{3}$

P.S

2. Originally Posted by Paymemoney
Hi
Can someone look where i have made a mistake because it doesn't match the answers in the book.

1) $\int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})$

$[\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]$

$[\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]$

$\frac{-6a^\frac{3}{2}}{3}$

the book's answer is $\frac{-5a^\frac{3}{2}}{3}$

2) $\int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}$

$6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}$

final answer is: $\frac{-38}{3}$
but book answer is $\frac{-55}{3}$

P.S
1. I suppose you were integrating with respect to x. Note that a is a constant, independent of x, so:

$\int_{a}^{4a} a^{\frac{1}{2}}- x^{\frac{1}{2}} ~ dx = a^{\frac{1}{2}}\int_{a}^{4a} 1 ~ dx - \int_{a}^{4a} x^{\frac{1}{2}}~dx =$ $a^{\frac{1}{2}}(4a-a) - (\frac{2}{3}(4^{\frac{3}{2}}a^{\frac{3}{2}}) - \frac{2}{3} a^{\frac{3}{2}} = \frac{9}{3}a^{\frac{3}{2}} - \frac{14}{3}a^{\frac{3}{2}} = \frac{-5}{3}a^{\frac{3}{2}}$

2. Your integral is correct, I guess you miscalculated.

$6\sqrt{4}-\frac{10 \cdot 4^{\frac{3}{2}}}{3} + \frac{2}{\sqrt{4}} = 6\cdot 2 - \frac{80}{3} + 1 = \frac{-41}{3}$

$6\sqrt{1} - \frac{10}{3} + \frac{2}{\sqrt{1}} = \frac{18}{3} - \frac{10}{3} + \frac{6}{3} = \frac{14}{3}$

$\frac{-41}{3} - \frac{14}{3} = \frac{-55}{3}$

3. Originally Posted by Paymemoney
Hi
Can someone look where i have made a mistake because it doesn't match the answers in the book.

1) $\int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})$

You must ALWAYS add to the integral $dx\,\,\,or\,\,\,dy$ or whatever, so that all can know with respect to what you're integrating. Here, I assume you're integrating wrt x (i.e., it must be $\int\limits_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})\,dx$ ), and thus the following step is WRONG because you did NOT take $a^{1\slash 2}$ as a constant, as you should...

$[\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]$

If you're indeed integrating wrt x, here it must be $\left[a^{1\slash 2}x-\frac{2}{3}x^{3\slash 2}\right]_a^{4a}$ $=3a\cdot a^{1\slash 2}-\frac{2}{3}(8a^{3\slash 2}-a^{3\slash 2})$ $=-\frac{5}{3}a^{3\slash 2}$ , as written in your book.

$[\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]$

$\frac{-6a^\frac{3}{2}}{3}$

the book's answer is $\frac{-5a^\frac{3}{2}}{3}$

2) $\int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}$

$6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}$

final answer is: $\frac{-38}{3}$
but book answer is $\frac{-55}{3}$

$6(\sqrt{4}-\sqrt{1})-\frac{10}{3}(4^{3\slash 2}-1)+2\left(\frac{1}{\sqrt{4}}-1\right)=6\cdot 1-\frac{10}{3}\cdot 7-1=5-\frac{70}{3}=-\frac{55}{3}$ , so your book = 2, you = 0.

Tonio

P.S
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4. Originally Posted by tonio
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thanks for the help