Q. Evaluate ∫36x−22x2−9x+12dx. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
Perform Polynomial Division: First, we need to perform polynomial long division or synthetic division to simplify the integrand (2x2−9x+12)/(x−2) since the numerator is a polynomial of higher degree than the denominator.
Integrate Simplified Function: Performing the polynomial long division, we divide 2x2−9x+12 by x−2:2x−5 ___________x - 2 | 2x2−9x+12 - (2x2−4x) ___________−5x+12 - (−5x+10) ___________2The result of the division is 2x−5 with a remainder of 2. So, the integrand can be rewritten as:x−20
Evaluate Definite Integral: Now we can integrate the function term by term:∫(2x−5+x−22)dx=∫2xdx−∫5dx+∫x−22dx
Substitute Values: Integrating each term gives us:∫2xdx=x2∫5dx=5x∫x−22dx=2ln∣x−2∣So the indefinite integral is:x2−5x+2ln∣x−2∣+C
Simplify Final Answer: Now we need to evaluate the definite integral from x=3 to x=6. We do this by substituting the upper and lower limits into the indefinite integral and finding the difference: F(6)−F(3)=[62−5(6)+2ln∣6−2∣]−[32−5(3)+2ln∣3−2∣]
Simplify Final Answer: Now we need to evaluate the definite integral from x=3 to x=6. We do this by substituting the upper and lower limits into the indefinite integral and finding the difference: F(6)−F(3)=[62−5(6)+2ln∣6−2∣]−[32−5(3)+2ln∣3−2∣] Substitute the values and simplify: F(6)−F(3)=[36−30+2ln4]−[9−15+2ln1] = [6+2ln4]−[−6+0] = 6+2ln4+6 = 12+2ln4
Simplify Final Answer: Now we need to evaluate the definite integral from x=3 to x=6. We do this by substituting the upper and lower limits into the indefinite integral and finding the difference: F(6)−F(3)=[62−5(6)+2ln∣6−2∣]−[32−5(3)+2ln∣3−2∣] Substitute the values and simplify: F(6)−F(3)=[36−30+2ln4]−[9−15+2ln1] = [6+2ln4]−[−6+0] = 6+2ln4+6 = 12+2ln4 Since ln1 is 0, we can simplify further: 12+2ln4=12+2ln(22)=12+4ln2
Simplify Final Answer: Now we need to evaluate the definite integral from x=3 to x=6. We do this by substituting the upper and lower limits into the indefinite integral and finding the difference: F(6)−F(3)=[62−5(6)+2ln∣6−2∣]−[32−5(3)+2ln∣3−2∣] Substitute the values and simplify: F(6)−F(3)=[36−30+2ln4]−[9−15+2ln1] = [6+2ln4]−[−6+0] = 6+2ln4+6 = 12+2ln4 Since ln1 is 0, we can simplify further: 12+2ln4=12+2ln(22)=12+4ln2 The final answer in simplest form with all logs condensed into a single logarithm is: x=60
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