Question

Let \( f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} \). If \( f(1) + f'(1) = \alpha e - {1 \over 6} \), then the value of 150\( \alpha \) is equal to ___________.

Hint:

Substitute \(u=x^3\); the integrand becomes \(\tfrac13\,e^u\,\dfrac{u^2}{(u^2+2u+2)^2}\), which is the derivative of \(\dfrac13\cdot\dfrac{e^u}{u^2+2u+2}\). Also \(f'(t)\) is the integrand at \(t\).

Answer 1

Correct Answer: 16

Step 1: Put \(u=x^3\), \(du=3x^2dx\): \(x^8dx=u^2\cdot\tfrac{du}3\). So \(f(t)=\tfrac13\int_0^{t^3}\dfrac{e^u u^2}{(u^2+2u+2)^2}du\).
Step 2: With \(D=u^2+2u+2\): \(\dfrac{d}{du}\left(\dfrac{e^u}{D}\right)=\dfrac{e^u(D-(2u+2))}{D^2}=\dfrac{e^uu^2}{D^2}\).
Step 3: \(f(1)=\tfrac13\left[\dfrac{e^u}{D}\right]_0^1=\tfrac13\left(\tfrac e5-\tfrac12\right)=\tfrac e{15}-\tfrac16\).
Step 4: \(f'(1)=e^{1}\cdot\dfrac{1}{(1+2+2)^2}=\dfrac e{25}\).
Step 5: \(f(1)+f'(1)=e\left(\tfrac1{15}+\tfrac1{25}\right)-\tfrac16=\tfrac{8e}{75}-\tfrac16\Rightarrow\alpha=\tfrac8{75}\).
Step 6: \(150\alpha=150\cdot\tfrac8{75}=16\).
Answer: 16