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JEE Advanced 2022 Paper 2 · Mathematics · Indefinite and Definite Integrals · medium

Let varphi(t)=∫-t^t(7x^6+10x^4+9x^2/(1+e^x dx for t>0. The value of varphi(1) is

Numerical answer

Let φ(t)=∫−tt7x6+10x4+9x21+ex dx\varphi(t)=\displaystyle\int_{-t}^{t}\frac{7x^{6}+10x^{4}+9x^{2}}{1+e^{x}}\,dx for t>0t>0. The value of φ(1)\varphi(1) is
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Answer: 6

Write f(x)=7x6+10x4+9x2f(x)=7x^6+10x^4+9x^2, an even function. Let I=∫−11f(x)1+exdxI=\int_{-1}^{1}\frac{f(x)}{1+e^{x}}dx. Substituting x→−xx\to-x gives I=∫−11f(x)ex1+exdxI=\int_{-1}^{1}\frac{f(x)e^{x}}{1+e^{x}}dx. Adding the two expressions, 2I=∫−11f(x) dx2I=\int_{-1}^{1}f(x)\,dx, so I=∫01f(x) dxI=\int_{0}^{1}f(x)\,dx because ff is even. Hence φ(1)=∫01(7x6+10x4+9x2) dx=1+2+3=6\varphi(1)=\int_0^1(7x^6+10x^4+9x^2)\,dx=1+2+3=6. In general φ(t)=t7+2t5+3t3\varphi(t)=t^7+2t^5+3t^3, which also satisfies φ(1)=6\varphi(1)=6. A numerical quadrature of the original integrand agrees with this value.
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