1
$\begingroup$

Find $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$

Choices:

A: 1
B: $\frac{1}{2}$
C: $\frac{1}{4}$
D: 0

My attempt is:

The question is directly equal to $$\int_0^1\left(e^{(1+x)^2} - \frac{3e^{(1+3x)}}{2\sqrt{1+3x}}\right)dx$$

$\endgroup$
  • $\begingroup$ I am new so don't misunderstand $\endgroup$ – Mark 7 Apr 6 at 4:28
  • 3
    $\begingroup$ I suppose that in the integral it is $e^{(1\color{red}{+}x)^2}$ $\endgroup$ – Claude Leibovici Apr 6 at 5:16
  • $\begingroup$ Hint: split into two integrals and make an obvious change of variable in the second integral. $\endgroup$ – zhoraster Apr 6 at 5:21
2
$\begingroup$

Let $u=\sqrt{1+3x}$, then $\frac{du}{dx}=\frac{3}{2u}$

\begin{align} \frac32 \int_0^1 \frac{\exp(1+3x)}{\sqrt{1+3x}} \, dx&= \frac32 \int_1^2\frac{\exp(u^2)}{u}\cdot \frac{2u}{3}\, du = \int_1^2 \exp(u^2)\,du \end{align}

Also, let $v=1+x$,

$$\int_0^1 \exp((1+x)^2) \, dx = \int_1^2 \exp(v^2) \, dv$$

Hence, the two integral cancels out.

$\endgroup$
  • $\begingroup$ Gotcha thank you $\endgroup$ – Mark 7 Apr 6 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.