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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 9 maths textbook solution

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Answer: It depends only on d.

Hint: Use \int x^{n} d x

Given: \int_{-2}^{2}\left(a x^{5}+b x^{3}+c x+d\right) d x


\mathrm{I}=\int_{-2}^{2}\left(a x^{5}+b x^{3}+c x+d\right) d x

    =\left[\frac{a x^{6}}{6}+\frac{b x^{4}}{4}+\frac{c x^{2}}{2}+d x\right]_{-2}^{2}


    \begin{aligned} &=\frac{a(2)^{6}}{6}+\frac{b(2)^{4}}{4}+\frac{c(2)^{2}}{2}+2 d-\frac{a(2)^{6}}{6}-\frac{b(2)^{4}}{4}-\frac{c(2)^{2}}{2}+2 d \\\\ &=4 \mathrm{~d} \end{aligned}

Thus, it depends only on d.


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