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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Integrate the following with respect to x: x5ex2

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प्रश्न

Integrate the following with respect to x:

x5ex2

बेरीज
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उत्तर

`int x^5 "e"^(x^2) "d"x = int (x2)^2 * "e"^(x^2) * x  "d"x`

Put x2 = t

2x dx = dt

⇒ x dx = `1/2  "dt"`

`int x^5 "e"^(x^2)  "d"x = int "t"^2 * "e"^"t" * 1/2  "dt"`

= `1/2 int "t"^2 "e"^"t"  "dt"`  .........(1)

Consider `int "t"^2  "e"^"t"  "dt"`

u = t2

u' = 2t

u" = 2

u"' = 0

dv = et dt

⇒ v = `int "e"^"t"  "dt"`

⇒ v = et

v1 = `int "v"  "dt"`

= `int "e"^"t" "dt"`

= et

v2 = `int "v"_1  "dt"`

= `int "e"^"t" "dt"`

= et 

v3 = `int "v"_2  "dt"`

= `int "e"^"t" "dt"`

= et 

`int u"  "dv"` = uv – u'v1 + u"v2 – u"'v3 + u"'v4 – ...........

`int "t"^2 "e"^"t"  "dt"` = t2 et – 2t et + 2et + 0et

= `("t"^2 - 2"t" + 2) "e"^"t" + "c"`

Substtuting in equation (1) we get

`int x^5  "e"^(x^2)  "d"x = 1/2["t"^2 - 2"t" + 2] "e"^"t" + "c"`

= `1/2 [(x^2)^2 - 2x^2 + 2] "e"^(x^2) + "c"`

`int x^5  "e"^(x^2)  "d"x =  1/2 [x^4 - 2x^2 + 2] "e"^(x^2) + "c"`

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

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 11 Integral Calculus
Exercise 11.7 | Q 3. (ii) | पृष्ठ २१०

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