Answered

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Evaluate the integral:
(31-x)/(x^2+x-20)dx from 0 to 1

Plz show all work

Sagot :

LammettHash

Split up the integrand into partial fractions:

(31 - x)/(x² + x - 20) = (31 - x)/((x + 5)(x - 4))

= a/(x + 5) + b/(x - 4)

= (a (x - 4) + b (x + 5))/((x + 5)(x - 4))

= (5b - 4a + (a + b) x)/((x + 5)(x - 4))

Then a + b = -1 and 5b - 4a = 31. Solving for a and b yields a = -4 and b = 3, so

(31 - x)/(x² + x - 20) = -4/(x + 5) + 3/(x - 4)

Now integrating is trivial; the antiderivative is

-4 ln|x + 5| + 3 ln|x - 4| + C

and by the fundamental theorem of calculus, we end up with

[tex]\displaystyle \int_0^1 \frac{31-x}{x^2+x-20} \, dx = (-4\ln|1+5|+3\ln|1-4|) - (-4\ln|0+5|+3\ln|0-4|) \\\\ = -4\ln(6) + 3\ln(3) + 4\ln(5) - 3\ln(4) \\\\ = \ln\left(\frac{3^3\cdot5^4}{4^3\cdot6^4}\right) \\\\ = \ln\left(\frac{625}{3072}\right) \approx -1.592[/tex]

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