For x in (0, 5π/2), define f(x) = ∫₀ˣ √t sin t dt. Then f has (A) local minimum at π and 2π (B) loca

Q: Why can't we just integrate $\sqrt{t} \sin t$ first?

Excellent question! While integration is a powerful tool, it is not always the most efficient path. In fact, $\int \sqrt{t} \sin t dt$ does not have a simple elementary antiderivative. We don't need the function itself; we only need its *rate of change* to find maxima and minima. The Leibniz Rule allows us to bypass the integration entirely!

Q: Why do we reject $x=0$ as a critical point?

Look at the domain! The problem specifies $x \in (0, 5\pi/2)$. The function is defined on an open interval starting strictly after zero. Even if the derivative were zero at $x=0$, it wouldn't be a local extremum within our domain. Always respect the boundaries of your playground!

Q: How do I know if it's a maximum or a minimum without the second derivative test?

You can use the First Derivative Test! Look at the sign of $f'(x) = \sqrt{x} \sin x$. If the derivative changes from positive to negative, you have a maximum. If it changes from negative to positive, you have a minimum. The second derivative test is just a faster, more algebraic way to confirm this behavior.

Question

For $x \in (0, 5\pi/2)$, define $f(x) = \int_0^x \sqrt{t} \sin t dt$. Then $f$ has

Options:

(A) local minimum at $\pi$ and $2\pi$

(B) local minimum at $\pi$ and local maximum at $2\pi$

(C) local maximum at $\pi$ and local minimum at $2\pi$

(D) local maximum at $\pi$ and $2\pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to unravel a problem that might look intimidating at first glance, but is actually a beautiful dance of calculus.

We are dealing with an accumulator function defined as:
$$f(x) = \int_0^x \sqrt{t} \sin t \, dt$$

Imagine this function as a machine that fills a bucket with water. As $x$ increases, the machine adds more water (area) to the bucket. Our goal is to find when the water level is at its highest (local maximum) and when it is at its lowest (local minimum) within the interval $x \in (0, 5\pi/2)$.

### The Power of Leibniz

When you see an integral with a variable upper limit, your first instinct might be to try and solve the integral. Stop! Take a breath.

We don't need to find the closed-form expression for the integral. We only need to know how it changes. This is where the **Leibniz Rule** comes to our rescue.

It tells us that the derivative of an integral with a variable upper limit is simply the integrand evaluated at that limit. Applying this to our function, we get:
$$f'(x) = \sqrt{x} \sin x$$

### Finding the Critical Points

Now that we have the derivative, we are in the driver's seat. To find the peaks and valleys, we set $f'(x) = 0$.

This gives us the equation:
$$\sqrt{x} \sin x = 0$$

Since our domain is $x \in (0, 5\pi/2)$, we know $x$ is never zero. Thus, the only way for this product to be zero is if $\sin x = 0$.

Within our specific interval $(0, 5\pi/2)$, the solutions are:
$$x = \pi \quad 	ext{and} \quad x = 2\pi$$
These are our critical points, the moments where the 'filling' of our bucket changes direction.

### The Second Derivative Test

While we could use the First Derivative Test, let's use the Second Derivative Test to be absolutely certain. We differentiate $f'(x) = \sqrt{x} \sin x$ using the product rule:
$$f''(x) = \frac{1}{2\sqrt{x}} \sin x + \sqrt{x} \cos x$$

Now, let's test our critical points. At $x = \pi$:
$$f''(\pi) = \frac{1}{2\sqrt{\pi}} \sin \pi + \sqrt{\pi} \cos \pi = 0 + \sqrt{\pi}(-1) = -\sqrt{\pi}$$

Because $f''(\pi) < 0$, the function is concave down, confirming a **local maximum at $x = \pi$**.

At $x = 2\pi$:
$$f''(2\pi) = \frac{1}{2\sqrt{2\pi}} \sin 2\pi + \sqrt{2\pi} \cos 2\pi = 0 + \sqrt{2\pi}(1) = \sqrt{2\pi}$$

Because $f''(2\pi) > 0$, the function is concave up, confirming a **local minimum at $x = 2\pi$**.

### Conclusion

We have successfully navigated the problem! We found a local maximum at $x = \pi$ and a local minimum at $x = 2\pi$.

Remember, calculus is not just about memorizing formulas; it is about understanding the behavior of functions. Keep exploring, keep questioning, and keep falling in love with the elegance of mathematics!

Animated Solution for Mathematics - Definite Integration: For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

Select Answer:

Visualized Solution

Defining the Accumulator Function $f (x)$

Applying the Leibniz Rule

Finding $f^{'} (x)$

Finding Critical Points

Solving $sin x = 0$

Visualizing the Derivative

The Second Derivative Test Setup

Calculating $f^{''} (x)$

Testing $x = π$

Conclusion for $x = π$

Testing $x = 2 π$

Conclusion for $x = 2 π$

Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Power of Leibniz

Finding the Critical Points

The Second Derivative Test

Conclusion

Similar Questions

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t (d t)$ for all $x \in R$ and $f : [0, 1/2] \to [0, \infty)$ be a continuous function. For $α \in [0, 1/2]$ , if $F^{'} (α) + 2$ is the area of the region bounded by $x = 0, y = 0, y = f (x)$ and $x = α$ , then $f (0)$ is

Let $f : R \to R$ be given by $f (x) = (x - 1) (x - 2) (x - 5)$ . Define $F (x) = \int_{0}^{x} f (t) d t, x > 0$ . Then which of the following options is/are correct ?

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

Let $f : R \to R$ be defined as $f (x) = e^{- x} sin x$ . If $F : [0, 1] \to R$ is a differentiable function such that $F (x) = \int_{0}^{x} f (t) d t$ , then the value of $\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$ lies in the interval

Let a function $f : [0, 5] \to R$ , be continuous, $f (1) = 3$ and $F$ be defined as: $F (x) = \int_{1}^{x} t^{2} g (t) d t$ , where $g (t) = \int_{1}^{t} f (u) d u$ . Then for the function $F$ , the point $x = 1$ is

Let $f (x) = \int_{1}^{x} 2 - t^{2} d t$ . Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(0,5π/2), define f(x)=∫0x​t​sintdt. Then f has

Select Answer:

Visualized Solution

Defining the Accumulator Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Applying the Leibniz Rule

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Finding Critical Points

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinx=0

Visualizing the Derivative

The Second Derivative Test Setup

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=π

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=π

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π