# Definite Integral

• Feb 24th 2009, 05:09 AM
Macleef
Definite Integral

Evaluate the given definite integral using the fundamental theorem of calculus. Round to 2 decimal places if needed.

$\int_{-4}^0 (2x + 6)^4\, dx$

$= \frac{1}{2} \frac{(2x + 6)^5}{5} |_0^{-4}$

$= [\frac{1}{10}(2(0) + 6)^5] - [\frac{1}{10}(2(-4) + 6)^5]$

$= 780.8$

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Find the area of the region R that lies under the given curve y = f(x) over the indicated interval $a \leq x \leq b$. Round to the 6th decimal places. Under $y = x^5e^{-x^6}$, over $0 \leq x \leq 1$.

$\int_1^0 x^5e^{-x^6}\, dx$

$u = x^6$
$du = 6x^5\, dx$

$= \frac{1}{6} \int_1^0 e^{-u}\, du$

$= [-\frac{1}{6}e^{-1}] - [-\frac{1}{6}e^0]$

$= 0.105353$

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A study indicates that t months from now the population of a town will be growing at a rate of $P'(t) = 8 + 2t^{\frac{2}{3}}$ people per month. By how much will the population growth of the town increase over the next 6 months? Round to the nearest integer.

$\int_6^0 8 + 2t^{\frac{2}{3}}\, dt$

$= 8t + \frac{6}{5}t^{\frac{5}{3}} |_6^0$

$= [8(6) + \frac{6}{5}(6)^{\frac{5}{3}}]$

$= 72$

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The number of bacteria present after t minutes of an experiment was $Q(t) = 2100e^{0.04t}$. What was the average number of bacteria present during the first 2 minutes of the experiment. Round to 1 decimal place.

$V = \frac{1}{2-0} \int_2^0 2100e^{0.04t}\, dt$

$= 1050 \int_2^0 e^{0.04t}\, dt$

$= 1050 (25e^{0.08} - 25e^{0})$

$= 2186.3$
• Feb 24th 2009, 05:48 AM
Jester
Quote:

Originally Posted by Macleef

Evaluate the given definite integral using the fundamental theorem of calculus. Round to 2 decimal places if needed.

$\int_0^{-4} (2x + 6)^4\, dx$

$= \frac{1}{2} \frac{(2x + 6)^5}{5} |_0^{-4}$

$= [\frac{1}{10}(2(0) + 6)^5] - [\frac{1}{10}(2(-4) + 6)^5]$

$= 780.8$

--------

Find the area of the region R that lies under the given curve y = f(x) over the indicated interval $a \leq x \leq b$. Round to the 6th decimal places. Under $y = x^5e^{-x^6}$, over $0 \leq x \leq 1$.

$\int_1^0 x^5e^{-x^6}\, dx$

$u = x^6$
$du = 6x^5\, dx$

$= \frac{1}{6} \int_1^0 e^{-u}\, du$

$= [-\frac{1}{6}e^{-1}] - [-\frac{1}{6}e^0]$

$= 0.105353$

---------------

A study indicates that t months from now the population of a town will be growing at a rate of $P'(t) = 8 + 2t^{\frac{2}{3}}$ people per month. By how much will the population growth of the town increase over the next 6 months? Round to the nearest integer.

$\int_6^0 8 + 2t^{\frac{2}{3}}\, dt$

$= 8t + \frac{6}{5}t^{\frac{5}{3}} |_6^0$

$= [8(6) + \frac{6}{5}(6)^{\frac{5}{3}}]$

$= 72$

------------

The number of bacteria present after t minutes of an experiment was $Q(t) = 2100e^{0.04t}$. What was the average number of bacteria present during the first 2 minutes of the experiment. Round to 1 decimal place.

$V = \frac{1}{2-0} \int_2^0 2100e^{0.04t}\, dt$

$= 1050 \int_2^0 e^{0.04t}\, dt$

$= 1050 (25e^{0.08} - 25e^{0})$

$= 2186.3$

Typically we hve

$\int_a^b f(x)\,dx$ where $a < b$. Second, if the antiderivative of $f(x) = F'(x)$ then the fundemental theorem of calculus is

$\int_a^b f(x)\,dx = \left. F(x)\right|_a^b = F(b) - F(a)$

From what I see above, you have two problems but they seem to self correct.
• Feb 24th 2009, 05:55 AM
Macleef
Can you point to those problems? I don't know where I went wrong...
• Feb 24th 2009, 06:21 AM
Jester
Quote:

Originally Posted by Macleef
Can you point to those problems? I don't know where I went wrong...

Let's look at the first one

Quote:

Originally Posted by Macleef

Evaluate the given definite integral using the fundamental theorem of calculus. Round to 2 decimal places if needed.

$\int_0^{-4} (2x + 6)^4\, dx$

$= \frac{1}{2} \frac{(2x + 6)^5}{5} |_0^{-4}$

$= [\frac{1}{10}(2(0) + 6)^5] - [\frac{1}{10}(2(-4) + 6)^5]$

$= 780.8$

Here, what you should have is

$\int_{-4}^0 (2x + 6)^4\, dx$

$= \frac{1}{2} \frac{(2x + 6)^5}{5} |_{-4}^0$

$= [\frac{1}{10}(2(-4) + 6)^5- [\frac{1}{10}(2(0) + 6)^5] = 780.8$

See the difference?
• Feb 24th 2009, 06:29 AM
Macleef
yes I do... I think... why is it -4 - 0? I thought if $\int_b^a$ then a - b, and in that problem, it's $\int_{-4}^{0}$, so wouldn't it be 0 - (-4)?

nvrm.. I wrote the problem wrong. It's suppose to be $\int_{-4}^{0}$, not $\int_{0}^{-4}$,