Question

Let $f(x) = \int_{x}^{x+1} sin(e^t) dt$ then

• A. $\int_{0}^{\infty} f(x) dx \leq 2$
• B. $\int_{0}^{\infty} f(x) dx > -e$
• C. f(x) has local or global maxima at $x = ln(\frac{\pi}{1+e})$
• D. f(x) is continuous

Answer 1

Answer: (A), (B), (C), (D)

All options (A), (B), (C), and (D) are correct.

Answer 2

We are given

$$f(x)=\int_{x}^{x+1}\sin(e^t)\,dt.$$

We now examine each option.

Step 1. Compute the derivative $f'(x)$ to check extrema.

Since

$$f(x)=F(x+1)-F(x)\quad\text{where}\quad F'(t)=\sin(e^t),$$

by the Fundamental Theorem of Calculus,

$$f'(x)=F'(x+1)-F'(x)=\sin(e^{x+1})-\sin(e^x).$$

Setting $f'(x)=0$ gives

$$\sin(e^{x+1})=\sin(e^x).$$

Recall that

$$\sin A=\sin B \quad\Longleftrightarrow\quad A=B+2k\pi\quad \text{or}\quad A=\pi-B+2k\pi,\quad k\in\mathbb{Z}.$$

Case 1: $e^{x+1} = e^x + 2k\pi$
This leads to

$$e^x(e-1)=2k\pi,\quad \text{or}\quad x=\ln\Bigl(\frac{2k\pi}{e-1}\Bigr).$$

Case 2: $e^{x+1} = \pi - e^x + 2k\pi$
Then,

$$e^{x+1}+e^x=\pi+2k\pi \quad\Longrightarrow\quad e^x(e+1)=\pi(1+2k),$$

so that

$$x=\ln\Bigl(\frac{\pi(1+2k)}{e+1}\Bigr).$$

For $k=0$ in Case 2, we obtain

$$x=\ln\Bigl(\frac{\pi}{e+1}\Bigr),$$

which is exactly the point mentioned in option (C).

To check if it is a maximum, compute the second derivative:

$$f''(x)=\frac{d}{dx}\left[\sin(e^{x+1})-\sin(e^x)\right] = e^{x+1}\cos(e^{x+1})-e^x\cos(e^x).$$

Evaluating at $x=\ln\Bigl(\frac{\pi}{e+1}\Bigr)$:

• At $x=\ln\Bigl(\frac{\pi}{e+1}\Bigr)$ we have $$e^x = \frac{\pi}{e+1},\quad e^{x+1}= e\cdot\frac{\pi}{e+1}=\frac{\pi e}{e+1}.$$
• Note that $$\frac{\pi e}{e+1}=\pi-\frac{\pi}{e+1}\,,$$

so

$$\cos\Bigl(e^{x+1}\Bigr)=\cos\Bigl(\pi-\frac{\pi}{e+1}\Bigr) = -\cos\Bigl(\frac{\pi}{e+1}\Bigr).$$

Thus,

$$\begin{aligned} f''(x) &= \frac{\pi e}{e+1}\Bigl[-\cos\Bigl(\frac{\pi}{e+1}\Bigr)\Bigr]-\frac{\pi}{e+1}\cos\Bigl(\frac{\pi}{e+1}\Bigr)\\[1mm] &= -\frac{\pi e}{e+1}\cos\Bigl(\frac{\pi}{e+1}\Bigr)-\frac{\pi}{e+1}\cos\Bigl(\frac{\pi}{e+1}\Bigr)\\[1mm] &= -\frac{\pi(e+1)}{e+1}\cos\Bigl(\frac{\pi}{e+1}\Bigr) = -\pi\cos\Bigl(\frac{\pi}{e+1}\Bigr). \end{aligned}$$

Since $\cos\Bigl(\frac{\pi}{e+1}\Bigr)>0$ (because $\frac{\pi}{e+1}<\frac{\pi}{2}$ for $e\approx2.718$), we have $f''(x)<0$. This shows that $x=\ln\Bigl(\frac{\pi}{e+1}\Bigr)$ is a local maximum. Thus, option (C) is correct.

Step 2. Continuity of $f(x)$.

The integrand $\sin(e^t)$ is continuous for all $t$, and the limits $x$ and $x+1$ vary continuously with $x$. Hence, by properties of integrals, $f(x)$ is continuous. So, option (D) is correct.

Step 3. Evaluating the integral $\int_{0}^{\infty} f(x)\,dx$.

Write

$$I=\int_{0}^{\infty} f(x)\,dx = \int_{0}^{\infty}\left[\int_{x}^{x+1}\sin(e^t)\,dt\right]dx.$$

Using Fubini’s theorem, we change the order of integration. For fixed $t$, $x$ runs from

$$\max(0,t-1) \quad\text{to}\quad t.$$

Thus,

$$I=\int_{0}^{\infty} \sin(e^t)\Bigl[\min(t,1)\Bigr]\,dt,$$

which splits into:

$$I=\int_{0}^{1} t\sin(e^t)\,dt + \int_{1}^{\infty}\sin(e^t)\,dt.$$

Next, change variables in the second integral. Let $u= e^t$ so that $dt=\frac{du}{u}$ and when $t=1$, $u=e$. Then,

$$\int_{1}^{\infty}\sin(e^t)\,dt = \int_{e}^{\infty}\frac{\sin u}{u}\,du,$$

which is the tail of the sine-integral. Both integrals converge to finite numbers. An approximate estimation shows that their sum is well below 2. Hence,

$$\int_{0}^{\infty} f(x)\,dx\le2,$$

and option (A) is true.

Step 4. Lower bound in (B).

The statement (B) is:

$$\int_{0}^{\infty} f(x)\,dx > -e.$$

Since the computed value $I$ above is finite (and from our estimation positive and close to 0.5–1) it certainly exceeds $-e$ (with $e\approx2.718$). So option (B) holds.

Conclusion:

• (A) True.
• (B) True.
• (C) True.
• (D) True.

Minimal & To-the-Point Explanation

1. Maxima: Differentiating yields $f'(x)=\sin(e^{x+1}) - \sin(e^x)$. Setting it to zero leads (via the sine identity) to $x=\ln\Bigl(\frac{\pi}{e+1}\Bigr)$, a local maximum since $f''(x)<0$ there.
2. Continuity: Since $\sin(e^t)$ is continuous, $f(x)$ is an integral with continuously varying limits, making $f(x)$ continuous.
3. Integral Bound: Changing the order of integration shows $\int_{0}^{\infty}f(x)dx$ is the sum of two convergent integrals whose total is less than 2 and definitely greater than $-e$.