Let the definite integral be defined by the formula ∫(a to b) f(x) dx = (b-a)/2 (f(a) + f(b)). For m

Q: Wait, why do we reject the negative root here? Or rather, why are we setting the derivative to zero?

Great question! We are looking for the point $c$ that maximizes the area. In calculus, the local extrema (maxima or minima) occur where the slope of the tangent to the function is zero. By setting $F'(c) = 0$, we are finding the stationary point of our area function. Since the problem states $f''(x) < 0$ (concave down), the second derivative test confirms this stationary point is indeed a maximum.

Q: Why does the term $f(c)$ disappear when we differentiate? It feels like it should stay.

It is a magical moment in algebra! When you differentiate the first term $\frac{c-a}{2}(f(a) + f(c))$, you get $\frac{1}{2}(f(a) + f(c)) + \frac{c-a}{2}f'(c)$. When you differentiate the second term $\frac{b-c}{2}(f(c) + f(b))$, you get $-\frac{1}{2}(f(c) + f(b)) + \frac{b-c}{2}f'(c)$. When you add these, the $\frac{1}{2}f(c)$ from the first term and the $-\frac{1}{2}f(c)$ from the second term cancel out perfectly. It is not a mistake; it is the beauty of the structure!

Q: Is this related to the Mean Value Theorem? The final answer looks exactly like the MVT formula.

You have a sharp eye! Yes, it is deeply connected. The result $f'(c) = \frac{f(b) - f(a)}{b-a}$ is precisely the statement of Lagrange's Mean Value Theorem. Geometrically, it means the area is maximized when the tangent at $c$ is parallel to the secant line connecting the endpoints $(a, f(a))$ and $(b, f(b))$. You have just derived a fundamental theorem of calculus through a geometric optimization problem!

Question

Passage:
Let the definite integral be defined by the formula $\int_a^b f(x) dx = \frac{b-a}{2} (f(a) + f(b))$. For more accurate result for $c \in (a, b)$, we can use $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx = F(c)$ so that for $c = \frac{a+b}{2}$, we get $\int_a^b f(x) dx = \frac{b-a}{4} (f(a) + f(b) + 2f(c))$.

Question 1:
$\int_0^{\pi/2} \sin x dx =$
(A) $\frac{\pi}{8}(1+\sqrt{2})$
(B) $\frac{\pi}{4}(1+\sqrt{2})$
(C) $\frac{\pi}{8\sqrt{2}}$
(D) $\frac{\pi}{4\sqrt{2}}$

Question 2:
If $\lim_{x 	o a} \frac{\int_a^x f(x) dx - \left(\frac{x-a}{2}ight)(f(x) + f(a))}{(x-a)^3} = 0$, then $f(x)$ is of maximum degree
(A) 4
(B) 3
(C) 2
(D) 1

Question 3:
If $f''(x) < 0 \forall x \in (a, b)$ and $c$ is a point such that $a < c < b$, and $(c, f(c))$ is the point lying on the curve for which $F(c)$ is maximum, then $f'(c)$ is equal to
(A) $\frac{f(b)-f(a)}{b-a}$
(B) $\frac{2(f(b)-f(a))}{b-a}$
(C) $\frac{2f(b)-f(a)}{2b-a}$
(D) 0

Sigma Percentile · Accepted Answer

### The Geometry of Approximation

Welcome, future engineers. Today, we are not just solving an integral; we are sculpting a solution. Imagine standing on a curve, a concave-down arc, and you are tasked with finding the 'sweet spot'—the point $c$ that maximizes the area of two trapezoids beneath it.

This is the essence of optimization. We start with the area function:

$$F(c) = \frac{c-a}{2}(f(a) + f(c)) + \frac{b-c}{2}(f(c) + f(b))$$

This function is our canvas. It represents the sum of two trapezoids: one on the interval $[a, c]$ and the other on $[c, b]$. The first trapezoid has a width of $(c-a)$ and heights $f(a)$ and $f(c)$, while the second has a width of $(b-c)$ and heights $f(c)$ and $f(b)$.

### The Calculus of Optimization

Now, we must find the maximum. To do this, we need to differentiate $F(c)$ with respect to $c$. This is where many students stumble, but let us take a breath.

We apply the product rule to both terms. For the first term, $\frac{c-a}{2}(f(a) + f(c))$, the derivative is:

$$\frac{1}{2}(f(a) + f(c)) + \frac{c-a}{2}f'(c)$$

For the second term, $\frac{b-c}{2}(f(c) + f(b))$, the derivative is:

$$-\frac{1}{2}(f(c) + f(b)) + \frac{b-c}{2}f'(c)$$

Notice the negative sign in the second term? That comes from the derivative of $(b-c)$ with respect to $c$, which is $-1$. This is a classic trap, but you are prepared for it.

### The Elegant Cancellation

Now, watch the magic happen. When we combine these derivatives, we get:

$$F'(c) = \frac{1}{2}(f(a) + f(c)) + \frac{c-a}{2}f'(c) - \frac{1}{2}(f(c) + f(b)) + \frac{b-c}{2}f'(c)$$

Look closely at the terms without $f'(c)$: $\frac{1}{2}f(a) + \frac{1}{2}f(c) - \frac{1}{2}f(c) - \frac{1}{2}f(b)$. The $f(c)$ terms cancel out completely, leaving us with $\frac{f(a) - f(b)}{2}$.

Now, look at the terms with $f'(c)$: $\frac{c-a}{2}f'(c) + \frac{b-c}{2}f'(c)$. Factoring out $\frac{f'(c)}{2}$, we get $(c-a + b-c) = b-a$.

So, the derivative simplifies to:

$$F'(c) = \frac{f(a) - f(b)}{2} + \frac{b-a}{2}f'(c)$$

This is the moment of clarity.

### The Grand Conclusion

To maximize the area, we set $F'(c) = 0$. This gives us:

$$\frac{f(a) - f(b)}{2} + \frac{b-a}{2}f'(c) = 0$$

Rearranging, we find:

$$\frac{b-a}{2}f'(c) = \frac{f(b) - f(a)}{2}$$

Multiplying by $2$ and dividing by $(b-a)$, we arrive at:

$$f'(c) = \frac{f(b) - f(a)}{b-a}$$

This is not just an algebraic result; it is a profound geometric truth. It tells us that the area is maximized when the tangent at $c$ is parallel to the secant line connecting the endpoints.

This is the **Lagrange Mean Value Theorem** in action! You have successfully navigated the calculus, avoided the traps, and arrived at a beautiful, fundamental result. Keep this intuition with you; it will serve you well in every challenge you face.

Animated Solution for Mathematics - Differentiation: Comprehension Passage

$\int_{0}^{π /2} sin x d x =$

Select Answer:

If $lim_{x \to a} \frac{\int _{a}^{x} f ( x ) d x - ( \frac{x - a}{2} ) ( f ( x ) + f ( a ))}{( x - a ) ^{3}} = 0$ , then $f (x)$ is of maximum degree

Select Answer:

If $f^{''} (x) < 0\forall x \in (a, b)$ and $c$ is a point such that $a < c < b$ , and $(c, f (c))$ is the point lying on the curve for which $F (c)$ is maximum, then $f^{'} (c)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Concave Curve and Trapezoids

Formulating the Total Area Function $F (c)$

Differentiating $F (c)$ with Respect to $c$

Applying the Product Rule to Each Term

Grouping and Simplifying Terms

Applying the First Derivative Test for Maxima

Solving for $f^{'} (c)$ and the Mean Value Theorem

The Sigma Insight: Maxima and Minima

The Geometry of Approximation

The Calculus of Optimization

The Elegant Cancellation

The Grand Conclusion

Similar Questions

If the total maximum value of the function $f (x) = (\frac{3 e}{2 s i n x})^{s i n^{2} x}, x \in (0, \frac{π}{2})$ , is $k$ , then $\left( rac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8$ is equal to

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Comprehension Passage

Which of the following is true?

Which of the following is true?

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

Let $g (x) = f (x) + f (1 - x)$ and $f^{''} (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $tan^{- 1} 2 α + tan^{- 1} (\frac{1}{α}) + tan^{- 1} (\frac{α + 1}{α})$ is equal to

Investigate for maxima and minima the function $f (x) = \int_{1}^{x} [2 (t - 1) (t - 2)^{3} + 3 (t - 1)^{2} (t - 2)^{2}] d t$

Let $f (x) = sin^{3} x + λ sin^{2} x, - π /2 < x < π /2$ . Find the intervals in which $λ$ should lie in order that $f (x)$ has exactly one minimum and exactly one maximum.

Let $f (x) = \frac{s i n π x}{x ^{2}}, x > 0$ . Let $x_{1} < x_{2} < x_{3} < \dots < x_{n} < \dots$ be all the points of local maximum of $f$ and $y_{1} < y_{2} < y_{3} < \dots < y_{n} < \dots$ be all the points of local minimum of $f$ . Then which of the following options is/are correct ?

Comprehension Passage

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

Match the statements/expressions in Column I with the open intervals in Column II.

List-I

List-II

Animated Solution for Mathematics - Differentiation: Comprehension Passage

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

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→a​(x−a)3∫ax​f(x)dx−(2x−a​)(f(x)+f(a))​=0, then f(x) is of maximum degree

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f′′(x)<0∀x∈(a,b) and c is a point such that a<c<b, and (c,f(c)) is the point lying on the curve for which F(c) is maximum, then f′(c) is equal to

Select Answer:

Visualized Solution

Visualizing the Concave Curve and Trapezoids

Formulating the Total Area Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(c)

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(c) with Respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Applying the Product Rule to Each Term

Grouping and Simplifying Terms

Applying the First Derivative Test for Maxima

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c) and the Mean Value Theorem

The Sigma Insight: Maxima and Minima

The Geometry of Approximation

The Calculus of Optimization

The Elegant Cancellation

The Grand Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the total maximum value of the function f(x)=(2sinx3​e​)sin2x,x∈(0,2π​), is k, then \left( rac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x2​et​(t−2)(t−3)dt for all x∈(0,∞), then

Comprehension Passage

Which of the following is true?

Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0ex​1+t2f′(t)​dt. Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Investigate for maxima and minima the function f(x)=∫1x​[2(t−1)(t−2)3+3(t−1)2(t−2)2]dt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=sin3x+λsin2x,−π/2<x<π/2. Find the intervals in which λ should lie in order that f(x) has exactly one minimum and exactly one maximum.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for 0<x<1?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function e−xf(x) assumes its minimum in the interval [0,1] at x=1/4, which of the following is true?

Match the statements/expressions in Column I with the open intervals in Column II.

List-I

List-II

$\int_{0}^{π /2} sin x d x =$

If $lim_{x \to a} \frac{\int _{a}^{x} f ( x ) d x - ( \frac{x - a}{2} ) ( f ( x ) + f ( a ))}{( x - a ) ^{3}} = 0$ , then $f (x)$ is of maximum degree

If $f^{''} (x) < 0\forall x \in (a, b)$ and $c$ is a point such that $a < c < b$ , and $(c, f (c))$ is the point lying on the curve for which $F (c)$ is maximum, then $f^{'} (c)$ is equal to

Formulating the Total Area Function $F (c)$

Differentiating $F (c)$ with Respect to $c$

Solving for $f^{'} (c)$ and the Mean Value Theorem

If the total maximum value of the function $f (x) = (\frac{3 e}{2 s i n x})^{s i n^{2} x}, x \in (0, \frac{π}{2})$ , is $k$ , then $\left( rac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8$ is equal to

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

Investigate for maxima and minima the function $f (x) = \int_{1}^{x} [2 (t - 1) (t - 2)^{3} + 3 (t - 1)^{2} (t - 2)^{2}] d t$

Let $f (x) = sin^{3} x + λ sin^{2} x, - π /2 < x < π /2$ . Find the intervals in which $λ$ should lie in order that $f (x)$ has exactly one minimum and exactly one maximum.

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?