By completing the following activity, Evaluate ∫12x+3x(x+2) dx Solution: Let I = ∫12x+3x(x+2) dx Let x+3x(x+2)=Ax+B(x+2) ∴ x + 3 = A(x + 2) + B. x ∴ A = □, B = □ ∴ I = ∫12[( )x+( )(x+2)]dx ∴ I = - Mathematics and Statistics

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Sum

By completing the following activity, Evaluate int_1^2 (x + 3)/(x(x + 2))  "d"x

Solution: Let I = int_1^2 (x + 3)/(x(x + 2))  "d"x

Let (x + 3)/(x(x + 2)) = "A"/x + "B"/((x + 2))

∴ x + 3 = A(x + 2) + B.x

∴ A = square, B = square

∴ I = int_1^2[("( )")/x + ("( )")/((x + 2))] "d"x

∴ I = [square log x + square log(x + 2)]_1^2

∴ I = square

Solution

Let I = int_1^2 (x + 3)/(x(x + 2))  "d"x

Let (x + 3)/(x(x + 2)) = "A"/x + "B"/((x + 2))

∴ x + 3 = A(x + 2) + B.x      ......(i)

Putting x = – 2 in (i), we get

– 2 + 3 = A(0) + B(– 2)

∴ 1 = – 2B

∴ B = -1/2

Putting x = 0 in (i), we get

0 + 3 = A(0 + 2) + B(0)

∴ 3 = 2A

∴ A = 3/2

∴ A = 3/2, B = -1/2

∴ I = int_1^2[(3/2)/x + ((-1/2))/(x + 2)] "d"x

∴ I = [3/2 log x + -1/2 log (x + 2)]_1^2

= 3/2 (log 2 - log 1) - 1/2 (log 4 - log 3)

= 3/2 (log 2 - 0) - 1/2 log (4/3)

= 1/2 log 2^3 - 1/2 log (4/3)

= 1/2 (log8 - log  4/3)

= 1/2 log (8 xx 3/4)

∴ I = 1/2 log 6

Concept: Fundamental Theorem of Integral Calculus
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Chapter 1.6: Definite Integration - Q.6
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