The value of -022 x3-6 x2-9 x-5-x2-2 x-5 d x equals

The correct option is D None of the aboveMake the substitution x – 1= t. ∫_ 1^12t+1^3 6t+1^2 + 9t+1 5/t+1^2 2t+1 + 5dt ∫_ 1^12t^3 +3t/t^2+4dt = 0 Since, ...

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Standard XII

Mathematics

Area between Two Curves

The value of ...

Question

The value of $\int_{0}^{2} \frac{2 x^{3} - 6 x^{2} + 9 x - 5}{x^{2} - 2 x + 5} d x$ equals

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None of the above

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Solution

The correct option is D None of the above
Make the substitution x – 1= t.
$\int_{- 1}^{1} \frac{2 {t + 1}^{3} - 6 {t + 1}^{2} + 9 t + 1 - 5}{{t + 1}^{2} - 2 t + 1 + 5} d t$
$\int_{- 1}^{1} \frac{2 t^{3} + 3 t}{t^{2} + 4} d t$
= 0
Since, the above expression is an odd function. And limits are equal and opposite in sign.

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The value of ∫202x3−6x2+9x−5x2−2x+5dx equals

The correct option is D None of the aboveMake the substitution x – 1= t. ∫1−12t+13−6t+12+9t+1−5t+12−2t+1+5dt ∫1−12t3+3tt2+4dt = 0 Since, the above expression is an odd function. And limits are equal and opposite in sign.

The value of $\int_{0}^{2} \frac{2 x^{3} - 6 x^{2} + 9 x - 5}{x^{2} - 2 x + 5} d x$ equals