Let I1=∫13(x2+x+3)f(x3−2x2−5x+2020)dx and I2=∫13(4x2−3x−2)f(x3−2x2−5x+2020)dx. If the value of 2I13I2=ab where a and b are co-prime, then the value of (a+b) is

Consider I2−I1=∫13(3x2−4x−5)f(x3−2x2−5x+2020)dxPut x3−2x2−5x+2020=t∴(3x2−4x−5)dx=dt∴ I2−I1=∫20142014f(t)dt=0Hence, I2−I1∴ 2I13I2=23=ab⇒a+b=5

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Let $I_{1} = \int_{1}^{3} (x^{2} + x + 3) f (x^{3} - 2 x^{2} - 5 x + 2020) d x$ and $I_{2} = \int_{1}^{3} (4 x^{2} - 3 x - 2) f (x^{3} - 2 x^{2} - 5 x + 2020) d x .$ If the value of $\frac{2 I_{1}}{3 I_{2}} = \frac{a}{b}$ where a and b are co-prime, then the value of $(a + b)$ is

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answer is 5.

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Consider $I_{2} - I_{1} = \int_{1}^{3} (3 x^{2} - 4 x - 5) f (x^{3} - 2 x^{2} - 5 x + 2020) d x$

Put $x^{3} - 2 x^{2} - 5 x + 2020 = t$

$∴ (3 x^{2} - 4 x - 5) d x = d t$

$∴ I_{2} - I_{1} = \int_{2014}^{2014} f (t) d t = 0$

Hence, $I_{2} - I_{1}$

$∴ \frac{2 I_{1}}{3 I_{2}} = \frac{2}{3} = \frac{a}{b} \Rightarrow a + b = 5$

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