Question

The value of \( \displaystyle \int_{-1}^{1} x^2 e^{x^3} \, dx \), where \( [t] \) denotes the greatest integer \( \leq t \), is :

• A. (e + 1)/3
• B. 1/(3e)
• C. (e - 1)/(3e)
• D. (e + 1)/(3e)

Hint:

When you see a greatest integer function $[t]$ inside an integral, always split the limits at points where the expression inside the brackets becomes an integer. For $\int_{-1}^{1} [x^2] e^{x^3} dx$: \begin{itemize} \item On the interval $x \in (-1, 1)$, we have $0 \leq x^2<1$, which means $[x^2] = 0$. \item Since the integrand is zero almost everywhere on the interval, the integral value is simply {0}. \end{itemize}

Answer 1

The Correct Option is C

Step 1: Let $t = x^3$, then $dt = 3x^2 dx \Rightarrow x^2 dx = dt/3$.
Step 2: When $x=-1, t=-1$. When $x=1, t=1$.
Step 3: Integral $= \frac{1}{3} \int_{-1}^{1} e^t dt = \frac{1}{3} [e^t]_{-1}^{1} = \frac{1}{3} (e - e^{-1}) = \frac{e - 1/e}{3} = \frac{e^2 - 1}{3e}$. *(Note: If the brackets [ ] represent the greatest integer function, the result for $\int [x^2]e^{x^3}$ would be 0 as $[x^2]=0$ for $x \in (-1, 1)$. Standard calculus problems of this type usually use brackets to signify standard grouping.)*