हिंदी

Evaluate the following integrals: ∫_2^7 sqrtx/(sqrtx + sqrt(9 − x))dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following integrals:

 `int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))*dx`

Evaluate: `∫_2^7 sqrtx/(sqrtx + sqrt(9 − x))dx`

मूल्यांकन
Advertisements

उत्तर

Let I = `int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))dx`     ...[i]

= `int_2^7 sqrt(2 + 7 - x)/(sqrt(2 + 7 - x) + sqrt(9 - (2 + 7 - x)))dx`            `...[∵ int_a^b f(x)dx = int_a^b f(a + b - x) dx]`

∴ I = `int_2^7 sqrt(9 - x)/(sqrt(9 - x) + sqrt(x))dx`    ...[ii]

Adding [i] and [ii], we get

2I = `int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))dx + int_2^7 sqrt(9 - x)/(sqrt(9 - x) + sqrt(x))dx` 

= `int_2^7 (sqrt(x) + sqrt(9 - x))/(sqrt(x) + sqrt(9 - x)) dx`

= `int_2^7 1dx`

= `[x]_2^7`

∴ 2I = 7 – 2

∴ 2I = 5

∴ I = `(5)/(2)`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Definite Integration - EXERCISE 6.2 [पृष्ठ १४८]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 6 Definite Integration
EXERCISE 6.2 | Q 6) | पृष्ठ १४८
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×