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Integrate the following with respect to x: 27x2e3x - Mathematics

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प्रश्न

Integrate the following with respect to x:

27x2e3x

बेरीज

उत्तर

`int 27x^2"e"^(3x)  "d"x = 27 int x^2  "e"^(3x)  "d"x`

u = x2

u'= 2x

u" = 2

u"' = 0

v = `"e"^(3x) * "d"x`

⇒ v = `int "e"^(3x) * "d"x`

= `("e"^(3x))/3`

v1 = `int "v"  "d"x`

= `int ("e"^(3x))/3  "d"x`

= `1/3 xx ("e"^3x)/3`

= `("e"^(3x))/3^2`

v2 = `int "v"_1  "d"x`

= `int ("e"^(3x))/3^2  "d"x`

= `1/3^2 xx ("e"^3x)/3`

= `1/3^3  "e"^(3x)`

v3 = `int "v"_2  "d"x`

= `int 1/3^3  "e"^(3x)  "d"x`

= `1/3^4  "e"^(3x)`

`int  "u"  "dv"` = uv – u'v1 + u"v2 – u"'v3 + ..........

`27 int x^2  "e"^(3x)  "d"x = 27[x^2  "e"^(3x)/3 - 2x "e"^(3x)/3^2 + 2 xx "e"^(3x)/3^3 - 0 xx "e"^(3x)/3^4] + "c"`

= `9x^2 "e"^(3x) - 6x "e"^(3x) + 2"e"^(3x) + 0 + "c"`

`27int x^2 "e"^(3x)  "d"x = (9x^2 - 6x + 2)"e"^(3x) + "c"`

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पाठ 11: Integral Calculus - Exercise 11.7 [पृष्ठ २१०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 11 Integral Calculus
Exercise 11.7 | Q 2. (ii) | पृष्ठ २१०

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