The value of \(\int_{1}^{4}\displaystyle\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2(e^{2}-e)\)
Let \(t = \sqrt{x}\), so
\( \displaystyle dt = \frac{dx}{2\sqrt{x}},\quad\text{that is}\quad\frac{dx}{\sqrt{x}} = 2\,dt \)
When \(x = 1\), \(t = 1\); when \(x = 4\), \(t = 2\). Then
\( \displaystyle \int_{1}^{4}\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = 2\int_{1}^{2}e^{t}\,dt \)
\( \displaystyle = 2\left[e^{t}\right]_{1}^{2} = 2(e^{2}-e) \)
Why the other options are incorrect
Option (1) \(e^{2}-e\) — Incorrect. It leaves out the factor 2 from \(\frac{dx}{\sqrt{x}} = 2\,dt\).
Option (3) \(2e^{2}\) — Incorrect. It leaves out the value at the lower limit t = 1.
Option (4) \(\displaystyle\frac{e^{2}-e}{2}\) — Incorrect. It divides by 2 instead of multiplying by 2.