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Evaluate the following : ∫03x2(3-x)52⋅dx - Mathematics and Statistics

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प्रश्न

Evaluate the following : `int_0^3 x^2(3 - x)^(5/2)*dx`

योग

उत्तर

Let I = `int_0^3 x^2(3 - x)^(5/2)*dx`

We use the property `int_0^a f(x)*dx = int_0^a f(a - x)*dx`

Here, a = 3
Hence in I, changing x to 3 – x, we get

I = `int_0^3 (3 - x)^2 [3 - (3 - x)]^(5/2)*dx`

= `int_0^3 (9 - 6x + x^2)x^(5/2)*dx`

= `int_0^3 [9x^(5/2) - 6x^(7/2) + x^(9/2)]*dx`

= `9 int_0^3 x^(5/2)*dx - 6int_0^3 x^(7/2)*dx + int_0^3 x^(9/2)*dx`

= `9[(x^(7/2))/(7/2)]_0^3 - 6[(x^(9/2))/(9/2)]_0^3 + 9[(x^(11/2))/(11/2)]_0^3`

= `9[(2.3^(7/2))/7 - 0] - 6[(2.3^(9/2))/9 - 0] + [2/11*3^(11/2) - 0]`

= `(18)/(7) 3^(7/2) - (2.6)/9*3^(7/2)*3 + 2/11*3^(7/2)*3^2`

= `2(3)^(7/2)[9/7 - 2 + 9/11]`

= `2(3)^(7/2)[(99 - 154 + 63)/77]`

= `2(3)^(7/2) xx 8/77`

= `16/77(3)^(7/2)`.

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Fundamental Theorem of Integral Calculus
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Exercise 4.2 [पृष्ठ १७२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 4 Definite Integration
Exercise 4.2 | Q 3.05 | पृष्ठ १७२

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