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Chapter 3: Problem 73

Let \(F(x)=\int_{x}^{x^{2}+\frac{\pi}{6}} 2 \cos ^{2} t d t\) for all \(x \in R\) and \(f:\left[0, \frac{1}{2}\right] \rightarrow[0, \infty)\) be a continuous function. For \(a \in\left[0, \frac{1}{2}\right]\), if \(F^{\prime}(a)+2\) is the area of the region bounded by \(x=0, y=0, y=f(x)\) and \(x=a\), then \(f(0)\) is

Short Answer

Expert verified
\( f(0) = 0 \). The function value at zero is 0.

Step by step solution

01

Differentiate the Integral

To find the derivative of \( F(x) = \int_{x}^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt \), we use the Leibniz rule for differentiation under the integral sign. The rule states that if \( G(x) = \int_{u(x)}^{v(x)} g(t) dt \), then \( G'(x) = g(v(x))v'(x) - g(u(x))u'(x) \). In this case, \( u(x) = x \) and \( v(x) = x^2 + \frac{\pi}{6} \). Then, \( F'(x) = 2\cos^2(x^2 + \frac{\pi}{6}) \cdot 2x - 2\cos^2(x) \cdot 1 \). Simplifying gives: \[ F'(x) = 4x \cos^2(x^2 + \frac{\pi}{6}) - 2 \cos^2(x). \]
02

Relate to the Area Condition

According to the problem, \( F'(a) + 2 \) is equal to the area of the region bounded by \( x=0, \, y=0, \, y=f(x), \) and \( x=a \). The area of this region is given by the integral: \( \int_{0}^{a} f(x) \, dx \). Thus, we have:\[ 4a \cos^2(a^2 + \frac{\pi}{6}) - 2 \cos^2(a) + 2 = \int_0^a f(x) \, dx. \]
03

Evaluate at the Endpoint

Note that at \( a = 0 \), the integral \( \int_0^a f(x) \, dx \) becomes zero since the bounds are the same. Hence:\[ 4 \cdot 0 \cdot \cos^2(0) - 2 \cos^2(0) + 2 = 0. \] The equation simplifies to:\[ -2 \cos^2(0) + 2 = 0. \] Since \( \cos(0) = 1 \), we have \( -2 \cdot 1^2 + 2 = 0 \), which simplifies correctly.
04

Conclude on the Function Value

Since the area results in zero at \( a = 0 \), matching the condition shows that:\[ f(0) = 0. \] This is because the integrals of zero-width (having both limits the same) contribute nothing, and since the other results match up, \( f(x) \) must be zero at that point to combine with the results correctly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation under the integral sign
Differentiation under the integral sign allows us to differentiate an integral with variable limits of integration. This technique is particularly useful in Integral Calculus, where direct evaluation may be challenging. When we have an integration, such as \( F(x) = \int_{u(x)}^{v(x)} g(t) \, dt \), and need its derivative, we can use the Leibniz rule.The Leibniz rule states that the derivative of \( F(x) \) with respect to \( x \) is \( G'(x) = g(v(x))v'(x) - g(u(x))u'(x) \).
  • \( g(t) \) is the function being integrated.
  • \( v(x) \) and \( u(x) \) are the variable bounds of integration.
  • The derivatives \( v'(x) \) and \( u'(x) \) are the rates of change of these bounds with respect to \( x \).
This approach enables us to deal with the variable limits systematically, making it a powerful tool in solving complicated integrals.
Area bounded by curves
Computing the area bounded by curves is a frequent application of integral calculus. In this context, the problem involves calculating the area of a region bounded by a set of curves. Specifically, given the boundaries \( x = 0, y = 0, y = f(x), \) and \( x = a \), we aim to find the area of the region enclosed.In this exercise, the area is determined by the integral:\[ \int_{0}^{a} f(x) \, dx. \]
  • The integral calculates the area under the curve represented by \( y = f(x) \) from \( x = 0 \) to \( x = a \).
  • The limits of the integral, \( 0 \) and \( a \), define the region of interest along the x-axis.
  • Understanding how to set up and evaluate this integral is crucial for solving problems involving areas bounded by curves.
Thus, integral calculus enables us to determine the area representation of physical and abstract boundaries effectively.
Cosine function
The cosine function, often denoted as \( \cos(x) \), is a fundamental trigonometric function with crucial properties in calculus and beyond. It helps describe oscillating phenomena such as waves and circular motion. Here are some key characteristics of the cosine function:
  • The cosine function has a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) units.
  • The range of \( \cos(x) \) is \([-1, 1]\), capturing its peak and trough values.
  • \( \cos(0) = 1 \): This value is vital for simplifying calculations, as seen in this problem.
  • \( \cos^2(x) \): This notation represents squaring the cosine function, commonly used to simplify trigonometric integrals.
Understanding these features of \( \cos(x) \) is essential for correctly evaluating integrals and derivatives in problems involving trigonometric functions.
Leibniz rule
The Leibniz rule is a central theorem in calculus, particularly useful in solving problems involving variable limits of integration. Named after the famous mathematician Gottfried Wilhelm Leibniz, this rule provides a formula for differentiating an integral where the limits are not constants.The Leibniz rule is expressed as:\[ G'(x) = g(v(x))v'(x) - g(u(x))u'(x) \]Where:
  • \( g(t) \) is the integrand or the function we are integrating.
  • \( v(x) \) and \( u(x) \) are the upper and lower bounds of the integral, respectively.
  • \( v'(x) \) and \( u'(x) \) are derivatives of these bounds with respect to \( x \).
In application, it allows us to solve the derivative \( F'(x) \) by substituting and simplifying based on the known functions. This is especially beneficial when tackling integrals with variable integration limits, as it systematically accounts for contributions from both bounds.

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Most popular questions from this chapter

Calculate the area enclosed by the curve $$ 4 \leq x^{2}+y^{2} \leq 2(|x|+|y|) $$

3 points \(O(0,0), P\left(a, a^{2}\right), Q\left(-b, b^{2}\right)(a>0, b>0)\) are on the parabola \(y=x^{2} .\) Let \(S_{1}\) be the area bounded by the line \(P Q\) and the parabola and let \(S_{2}\) be the area of the \(\triangle O P Q\). the minimum value of \(S_{1} / S_{2}\) is (a) \(4 / 3\) (b) \(5 / 3\) (c) 2 (d) \(7 / 3\)

The triangle formed by the normal to the curve \(f(x)=x^{2}-a x+2 a\) at the point \((2,4)\) and the coordinate axes lies in second quadrant, if its area is 2 sq units, then a can be (a) 2 (b) \(17 / 4\) (c) 5 (d) \(19 / 4\)

Let \(f(t)=|t-1|-|t|+|t+1|, \forall t \in R\) and \(g(x)=\max \\{f(t): x+1 \leq t \leq x+2\\} ; \forall x \in R\). Find \(g(x)\) and the area bounded by the curve \(y=g(x)\), the \(X\) -axis and the lines \(x=-3 / 2\) and \(x=5\).

Consider \(f(x)=x^{2}-3 x+2\) The area bounded by \(|y|=|f(|x|)|, x \geq 1\) is \(A\), then find the value of \(3 A+2\)
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