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Definite Integration question

2022 · 30 Jun · Shift 1 · Q37
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  5. /2022 · 30 Jun · Shift 1 · Q37

Definite Integration question

2022 · 30 Jun · Shift 1 · Q37

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let f(t)=∫0tex3(x8(x6+2x3+2)2)dxf(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx}f(t)=0∫t​ex3((x6+2x3+2)2x8​)dx. If f(1)+f′(1)=αe−16f(1) + f'(1) = \alpha e - {1 \over 6}f(1)+f′(1)=αe−61​, then the value of 150 α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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