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Syllabus

Find: ∫ ( 3 X + 5 ) √ 5 + 4 X − 2 X 2 D X .

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Question

Find: `int_  (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.

Sum
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Solution

`( 3x + 5)sqrt(5 + 4x - 2x^2) dx`

Let 3x + 5 = A(4 - 4x) + B

⇒ A = `-3/4, B = 8`

I = `3/4(4 - 4x)sqrt(5 + 4x + 2x^2) dx + 8 sqrt(5 + 4x - 2x^2)dx`

= `-3/4 I_1 + 8l_2 ("let")`

For I1, put 5 + 4x = - 2x2 = t

⇒ (4 - 4x) dx = dt

`-3/4 I_1 = - 3/4 sqrtt dt = - 3/4 xx 2/3 t^(3/2)`

= `-1/2 (5 + 4x - 2x^2 )^(3/2)`

`8I_2 = 8sqrt2 sqrt(7/2 - ( x - 1)^2) dx`

I = `-1/2 (5 + 4x - 2x^2 )^(3/2) + 4sqrt2(x - 1) sqrt(5/2 + 2x - x^2) + 14sqrt2 sin^-1  (sqrt2(x - 1))/sqrt7 + C`

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Evaluation of Definite Integrals by Substitution
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Q 14
2015-2016 (March) All India Set 1 E

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2015-2016 (March) All India Set 1 E (with solutions)
Q 14 | 4 marks

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