If ∫01x1−1−x21002019dx∫01x1−1−x21002020dx=pqp,q,∈N, G.C.D of p,q=1 then sum of digits of p+q equals

GE=I1I2=∫011−t1002019dt∫011−t1002020dtI2=∫011−t1002020.1.dt=0−∫0120201−t10020191−t100−1dt by parts=−202000I2−I1∴I1I2=202001202000=pq=p+q=404001Sum of digits=9

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If $\frac{{\int_{0}^{1} x (1 - {(1 - x^{2})}^{100})}^{2019} dx}{\int_{0}^{1} x {(1 - {(1 - x^{2})}^{100})}^{2020} dx} = \frac{p}{q}$ $(p, q, \in N, G . C . D of (p, q) = 1)$
then sum of digits of $p + q$ equals

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answer is 9.

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Detailed Solution

GE= $\frac{I_{1}}{I_{2}} = \frac{\int_{0}^{1} {(1 - t^{100})}^{2019} dt}{\int_{0}^{1} {(1 - t^{100})}^{2020} dt}$
$I_{2} = \int_{0}^{1} {(1 - t^{100})}^{2020} . 1 . dt = 0 - \int_{0}^{1} 2020 {(1 - t^{100})}^{2019} (1 - t^{100} - 1) dt$ by parts
$= - 202000 (I_{2} - I_{1})$
$∴ \frac{I_{1}}{I_{2}} = \frac{202001}{202000} = \frac{p}{q} = p + q = 404001$
Sum of digits=9