मराठी

Evaluate the definite integral: ∫12(4x3-5x2+6x+9) dx

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प्रश्न

Evaluate the definite integral:

`int_1^2 (4x^3 - 5x^2 + 6x + 9)  dx`

बेरीज
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उत्तर

`= [(4x^4)/4 - (5x^3)/3 + (6x^2)/2 + 9x]_1^2`

`= (x^4 - 5/3  x^3  + 3x^2 + 9x)_1^2`

`= (2^4 - 1^4) - 5/3 (2^3 - 1^3) + 3 (2^2 - 1^2) + 9 (2 - 1)`

`= (16 - 1) - 5/3 (8 - 1) + 3 (4 - 1) + 9 (1)`

`= 15 - 35/3 + 9 + 9`

`= 33 - 35/3`

`= (99 - 35)/3`

`= 64/3`

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पाठ 7: Integrals - Exercise 7.9 [पृष्ठ ३३८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.9 | Q 3 | पृष्ठ ३३८

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