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सी. बी. एस. ई.वाणिज्य (इंग्रजी माध्यम) इयत्ता १२
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अभ्यासक्रम

Evaluate: ∫ π − π ( 1 − X 2 ) Sin X Cos 2 X D X .

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प्रश्न

Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x"  d"x"`.

बेरीज
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उत्तर

`int_-π^π (1 - "x"^2) sin "x"  cos^2 "x"  d"x"`

We know
`int_-a^a "f" ("x")"d" "x" = 0` if f is an odd function i.e i f  f (-x) = -f (x)

In the given integral,

`"f" ("x") = (1 - "x"^2) sin "x" cos^2 "x"`

⇒ `"f" (- "x") = (1- (-"x")^2) (sin (-"x")) cos^2 (-"x") = -(1 -"x"^2) sin "x" cos^2 "x"`

⇒ `"f" (-"x") = -"f" ("x")`

So, `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" "dx" = 0`

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Evaluation of Definite Integrals by Substitution
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
Q 6.1
2018-2019 (March) 65/1/3

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2018-2019 (March) 65/1/3 (with solutions)
Q 6.1 | 2 marks

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