Mathematics, 07.07.2020 19:01, Averybeam300

Evaluate the following definite integral.

10
∫ 13y/ y^2-9y-22 .dy
-1

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answered: ineedhelp2285

$\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = -9 \ln (12)$

General Formulas and Concepts:

Pre-Calculus

Partial Fraction Decomposition

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13\int\limits^{10}_{-1} {\frac{y}{y^2 - 9y - 22}} \, dy$ [Integrand] Factor: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13\int\limits^{10}_{-1} {\frac{y}{(y - 11)(y + 2)}} \, dy$

Step 3: integrate Pt. 2

[Integrand] Split [Partial Fraction Decomp]: $\displaystyle \frac{y}{(y - 11)(y + 2)} = \frac{A}{y - 11} + \frac{B}{y + 2}$ [Decomp] Rewrite: $\displaystyle y = A(y + 2) + B(y - 11)$ [Decomp] Substitute in y = -2: $\displaystyle -2 = A(-2 + 2) + B(-2 - 11)$ Simplify: $\displaystyle -2 = -13B$ Solve: $\displaystyle B = \frac{2}{13}$ [Decomp] Substitute in y = 11: $\displaystyle 11 = A(11 + 2) + B(11 - 11)$ Simplify: $\displaystyle 11 = 13A$ Solve: $\displaystyle A = \frac{11}{13}$ [Split Integrand] Substitute in variables: $\displaystyle \frac{y}{(y - 11)(y + 2)} = \frac{\frac{11}{13}}{y - 11} + \frac{\frac{2}{13}}{y + 2}$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Split Integrand]: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13\int\limits^{10}_{-1} {\bigg( \frac{\frac{11}{13}}{y - 11} + \frac{\frac{2}{13}}{y + 2} \bigg)} \, dy$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13 \bigg[ \int\limits^{10}_{-1} {\frac{\frac{11}{13}}{y - 11}} \, dy + \int\limits^{10}_{-1} {\frac{\frac{2}{13}}{y + 2}} \, dy \bigg]$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13 \bigg[ \frac{11}{13}\int\limits^{10}_{-1} {\frac{1}{y - 11}} \, dy + \frac{2}{13}\int\limits^{10}_{-1} {\frac{1}{y + 2}} \, dy \bigg]$

Step 5: Integrate Pt. 4

Identify variables for u-substitution.

Integral 1

Set u: $\displaystyle u = y - 11$ [u] Differentiation [Basic Power Rule, Derivative Properties]: $\displaystyle du = dy$ [Bounds] Switch: $\displaystyle \left \{ {{y = 10 ,\ u = 10 - 11 = -1} \atop {y = -1 ,\ u = -1 - 11 = -12}} \right.$

Integral 2

Set v: $\displaystyle v = y + 2$ [v] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle dv = dy$ [Bounds] Switch: $\displaystyle \left \{ {{y = 10 ,\ v = 10 + 2 = 12} \atop {y = -1 ,\ v = -1 + 2 = 1}} \right.$

Step 6: Integrate Pt. 5

[Integrals] U-Substitution: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13 \bigg[ \frac{11}{13}\int\limits^{-1}_{-12} {\frac{1}{u}} \, du + \frac{2}{13}\int\limits^{12}_{1} {\frac{1}{v}} \, dv \bigg]$ [Integrals] Logarithmic Integration: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13 \bigg[ \frac{11}{13}(\ln |u|) \bigg| \limits^{-1}_{-12} + \frac{2}{13}(\ln |v|) \bigg| \limits^{12}_{1} \bigg]$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 13 \bigg[ \frac{11}{13}[-\ln (12)] + \frac{2}{13}[\ln (12)] \bigg]$ Simplify: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = 11[-\ln (12)] + 2[\ln (12)]$ Simplify: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = -11\ln (12)] + 2\ln (12)$ Simplify: $\displaystyle \int\limits^{10}_{-1} {\frac{13y}{y^2 - 9y - 22}} \, dy = -9 \ln (12)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

answered: Guest

step-by-step explanation:

answered: Guest

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.