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DETAILS MY NOTES SESSCALCET2 6.3.017. Evaluate the integral. \[ \int_{1}^{2} \...

Jun 30, 2024

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Solution by Steps

step 1

To evaluate the integral, we start by decomposing the integrand into partial fractions. The integrand is

\frac{4y^2 - 7y - 12}{y(y+2)(y-3)}

step 2

We express

\frac{4y^2 - 7y - 12}{y(y+2)(y-3)}

\frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-3}

step 3

Multiplying both sides by

y(y+2)(y-3)

, we get

4y^2 - 7y - 12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)

step 4

To find

A

B

, and

C

, we solve the system of equations obtained by equating coefficients of like terms

step 5

Solving, we get

A = 1

B = 2

, and

C = -3

. Thus,

\frac{4y^2 - 7y - 12}{y(y+2)(y-3)} = \frac{1}{y} + \frac{2}{y+2} - \frac{3}{y-3}

step 6

Now, we integrate each term separately:

\int_{1}^{2} \left( \frac{1}{y} + \frac{2}{y+2} - \frac{3}{y-3} \right) dy

step 7

This gives us

\int_{1}^{2} \frac{1}{y} dy + \int_{1}^{2} \frac{2}{y+2} dy - \int_{1}^{2} \frac{3}{y-3} dy

step 8

Evaluating each integral, we get

\left[ \ln|y| \right]_{1}^{2} + 2\left[ \ln|y+2| \right]_{1}^{2} - 3\left[ \ln|y-3| \right]_{1}^{2}

step 9

Substituting the limits, we get

\ln(2) - \ln(1) + 2(\ln(4) - \ln(3)) - 3(\ln(-1) - \ln(-2))

step 10

Simplifying, we get

\ln(2) + 2\ln\left(\frac{4}{3}\right) - 3\ln\left(\frac{2}{1}\right)

step 11

This simplifies to

\ln(2) + 2\ln\left(\frac{4}{3}\right) - 3\ln(2)

step 12

Combining the logarithms, we get

\ln\left(\frac{2 \cdot (4/3)^2}{2^3}\right)

step 13

Simplifying further, we get

\ln\left(\frac{32}{27}\right)

step 14

Therefore, the value of the integral is

\ln\left(\frac{32}{27}\right) \approx 1.7655

Answer

\ln\left(\frac{32}{27}\right) \approx 1.7655

Key Concept

Partial Fraction Decomposition

Explanation

The integral was evaluated by decomposing the rational function into partial fractions and then integrating each term separately.