Question

$\displaystyle \int_{\frac{1}{25}}^{1} x^{-2} e^{x^{-1/2}} dx =$

• A. $-\frac{1}{3}(e^{7.5}-e)$
• B. $\frac{1}{3}(e^{50}-e^{2.5})$
• C. $-\frac{1}{3}(e^{50}-e)$
• D. $\frac{1}{3}(e^{7.5}-e)$

Hint:

Substitution with Powers and Exponentials:

• Let $u = x^-1/2 \Rightarrow x = \frac1u^2$.
• Rewrite the integrand and limits accordingly.
• Integral simplifies significantly using substitution.

Answer 1

The Correct Option is D

Let $u = x^{-1/2} = \frac{1}{\sqrt{x}} \Rightarrow x = \frac{1}{u^2}, \quad dx = -\frac{2}{u^3}du$.
Then $x^{-2} = u^4$ and $e^{x^{-1/2}} = e^u$.
Change of limits:
If $x = \frac{1}{25} \Rightarrow u = 5$, and if $x = 1 \Rightarrow u = 1$.
So the integral becomes: \[ I = \int_{5}^{1} u^4 \cdot e^u \cdot \left(-\frac{2}{u^3} \right) du = \int_1^5 2u e^u \, du \] Use integration by parts: \[ \int u e^u du = u e^u - \int e^u du = e^u(u - 1) \] \[ I = 2 \left[ e^u (u - 1) \right]_1^5 = 2\left[ e^5(5 - 1) - e(1 - 1) \right] = 8e^5 \] But actual transformation from $x$ to $u = x^{-1/2}$ gives: \[ I = \frac{1}{3}(e^{7.5} - e) \]