Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 months ago

cal 2 integral question?

Please help! I got the answer ln(5)- ln(4)+3ln(2).

I plugged in the question into an integral website and it got ln(5)-ln(4)-2ln(2). Don't know where I went wrong!

Attachment image

5 Answers

Relevance
  • 5 months ago
    Favorite Answer

    husoski got that right.

    I am going to check that, just as a mental gymnastics, in different order of operations.

    ₀∫¹(x-8)/(x²-7x+10) dx = 

    substitute immediately

    u=2x-7 (yes, that's derivative of the denominator), => x=(u+7)/2,  du=2dx

    new lower bound u=2*0-7=-7 upper bound u=2*1-7=-5

    = (1/2) ₋₇∫⁻⁵ ((u+7)/2-8)/((u+7)²/4-7(u+7)/2+10) du

    = (1/4) ₋₇∫⁻⁵ ((u+7)-16)/((u+7)²/4-7(u+7)/2+10) du

    = ₋₇∫⁻⁵((u+7)-16)/((u+7)²-14(u+7)+40) du =

    = ₋₇∫⁻⁵(u-9)/(u²+14u+49-14u-14*7+40) du =

    = ₋₇∫⁻⁵(u-9)/(u²-9) du =

    partial fractions

    = ₋₇∫⁻⁵(2/(u+3) - 1/(u-3)) du =

    integrate term by term

    = 2₋₇∫⁻⁵1/(u+3) du - ₋₇∫⁻⁵1/(u-3) du

    = 2ln|u+3|₋₇⁻⁵ - ln|u-3|₋₇⁻⁵

    = 2 [ln|-5+3| - ln|-7+3|] - [ln|-5-3| - ln|-7-3| ]

    = 2 [ln|-2| - ln|-4|] - [ln|-8| - ln|-10| ]

    = 2 [ln(2)-ln(4)] - [ln(8)-ln(10) ]

    = -1.16315...

    Edit:

    The result can be shortened using basic operations with logarythms

    2(ln(2)-ln(4)) - (ln(8)-ln(10)) =

    = 2 ln(2/4) - ln(8/10)

    = 2 ln(1/2) - ln(4/5)

    = ln((1/2)²) - ln(4/5)

    = ln((1/4)/(4/5))

    = ln(5/16)

  • 5 months ago

    You used partial fractions, right?  The denominator factors as (x - 2)(x - 5), so you want to find constants A, B so that:

        (x - 8) / (x² - 7x + 10) = A / (x - 2) + B / (x - 5)

    Multiply by (x² - 7x + 10) = (x - 2)(x - 5):

        x - 8  = (x - 5)A + (x - 2)B = (A + B)x - 5A - 2B

    The coefficients on x and the constant terms must be independently equal:

        1   =  A + B

        -8  = -5A - 2B

    Negate the 2nd eq. and substitute B = 1 - A (from the 1st eq.):

         8  =  5A + 2B =  5A + 2(1 - A)

         6  =  5A - 2A = 3A

         A  =  6/3 = 2

         B  =  1 - A  =  -1

         (x - 8) / (x² - 7x + 10)  =  2/(x - 2) - 1/(x - 5)

         ∫ (x - 8) / (x² - 7x + 10) dx  =  2 ln |x - 2| - ln |x - 5| + C

    The definite integral follows from that.

    I'd simplify the result, though:

        ln 5 - ln 4 - 2 ln 2 = ln [ 5 / (4 * 2²) ] = ln (5/16)

    ...or ln 5 - 4 ln 2 if you only want logs of small integers.

  • rotchm
    Lv 7
    5 months ago

    Apply what I already told you in:

    https://ca.answers.yahoo.com/question/index?qid=20...

    Anything in there you still need help with before we tackle the current question? 

    This is how you will *learn*!

  • Ash
    Lv 7
    5 months ago

    ₀∫¹ (x-8)/(x²-7x+10) dx

    = ₀∫¹ (x-8)/(x-5)(x-2) dx

    (x-8)/(x-5)(x-2) = A/(x-5) + B/(x-2)

    (x-8)/(x-5)(x-2) = [A(x-2) + B(x-5)]/(x-5)(x-2)

    (x-8) = (Ax - 2A + Bx - 5B) 

    (x-8) = (A+B)x - (2A + 5B)

    then we get

    A+B = 1   .....(1)   and

    2A+5B = 8 ...(2)

    Multiple (1) by 2 and subtract from (2)

    2A+5B - 2(A+B) = 8 - 2

    2A+5B - 2A - 2B = 6

    3B = 6

    B = 2

    plug in (1)

    A + 2 = 1

    A = -1

    Then the integration becomes

    = ₀∫¹ -1/(x-5) + 2/(x-2) dx

    = -₀∫¹1/(x-5) dx + 2₀∫¹1/(x-2) dx

     

    Now we know ∫1/(x+a) = 1/a ln|ax+b| + C , then we get

    = [-ln|x-5| + 2ln|x-2|] ₀|¹

    = (-ln|1-5| + 2ln|1-2|) -  (-ln|0-5| + 2ln|0-2|)

    = -ln4 + 2ln1 + ln5 - 2ln2

    = ln5 - ln4 - 2 ln2

    we can further simplify as

    = ln5 - ln2² - 2 ln2

    = ln5 - 2ln2 - 2ln2

    = ln5 - 4ln2

  • How do you think about the answers? You can sign in to vote the answer.
  • ted s
    Lv 7
    5 months ago

    carelessness ????ln | x - 5 | + 2 ln | x - 2 |  ===> - ln 5

Still have questions? Get your answers by asking now.