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Math Help - Definite Integral

  1. #1
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    Definite Integral

    Please check my answers. Thank you.

    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

    --------

    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
    Last edited by Macleef; February 24th 2009 at 06:30 AM.
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  2. #2
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    Quote Originally Posted by Macleef View Post
    Please check my answers. Thank you.

    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.
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  3. #3
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    Can you point to those problems? I don't know where I went wrong...
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  4. #4
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    Quote Originally Posted by Macleef View Post
    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 View Post
    Please check my answers. Thank you.

    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?
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  5. #5
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    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},
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