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Find the derivatives. a. by evaluating the integral and differentiating the result. b. by differentiating the integral directly. $$\frac{d}{d t} \int_{0}^{\sqrt{t}}\left(x^{4}+\frac{3}{\sqrt{1-x^{2}}}\right) d x$$

Short Answer

Expert verified
The derivative is \( \frac{t^{3/2}}{2} + \frac{3}{2 \sqrt{t(1-t)}} \).

Step by step solution

01

Evaluate the Integral

First, compute the indefinite integral of the function \( f(x) = x^4 + \frac{3}{\sqrt{1-x^2}} \). This involves integrating each term separately. The integral \( \int x^4 \, dx \) yields \( \frac{x^5}{5} + C \). The integral \( \int \frac{3}{\sqrt{1-x^2}} \, dx \) evaluates to \( 3 \arcsin(x) + C \). The indefinite integral of \( f(x) \) is therefore \( \frac{x^5}{5} + 3 \arcsin(x) + C \).
02

Compute the Definite Integral

Now, evaluate the definite integral \( \int_{0}^{\sqrt{t}} \left(x^4 + \frac{3}{\sqrt{1-x^2}}\right) dx \) using the result from Step 1. Substitute \( \sqrt{t} \) and 0 into the indefinite integral: \[ \left( \frac{(\sqrt{t})^5}{5} + 3 \arcsin(\sqrt{t}) \right) - \left( \frac{0^5}{5} + 3 \arcsin(0) \right). \] Simplifying gives \( \frac{t^{5/2}}{5} + 3 \arcsin(\sqrt{t}) \).
03

Differentiate the Resulting Expression

Differentiate \( \frac{t^{5/2}}{5} + 3 \arcsin(\sqrt{t}) \) with respect to \( t \). Using derivative rules: The derivative of \( \frac{t^{5/2}}{5} \) is \( \frac{5}{10} t^{3/2} = \frac{t^{3/2}}{2} \). The derivative of \( 3 \arcsin(\sqrt{t}) \) is found using the chain rule: \( \frac{3}{\sqrt{1-(\sqrt{t})^2}} \times \frac{1}{2\sqrt{t}} = \frac{3}{2 \sqrt{t(1-t)}} \). Thus, the derivative is \( \frac{t^{3/2}}{2} + \frac{3}{2 \sqrt{t(1-t)}} \).
04

Differentiate the Integral Directly - Apply Leibniz Rule

Differentiate \( \int_{0}^{\sqrt{t}} \left(x^4 + \frac{3}{\sqrt{1-x^2}}\right) dx \) directly with respect to \( t \) using the Fundamental Theorem of Calculus. The theorem gives \( \frac{d}{dt} \int_{a(t)}^{b(t)} f(x) \, dx = f(b(t))b'(t) - f(a(t))a'(t) \). Evaluate at \( x=\sqrt{t} \) and \( x=0 \) (only \( x=\sqrt{t} \) contributes as \( a(t) = 0 \) gives no added term): \[ \left((t^2 + \frac{3}{\sqrt{1-t}})\right) \times \frac{1}{2\sqrt{t}} = \frac{t^2 + \frac{3}{\sqrt{1-t}}}{2\sqrt{t}}. \] Simplify to match the derivative obtained previously.

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone concept in calculus that links the concept of differentiation with that of integration. It has two main parts, but here we focus on how it helps to differentiate an integral over a function of a variable.

  • Part 1: It states that if you have a continuous function on an interval, the function's integral over that interval can be differentiated to obtain the original function.
  • Part 2: It tells us that the derivative of an integral from a to x of f(t) dt is simply f(x).
In practice, this theorem provides a powerful tool for differentiating integrals with variable limits, like in our example exercise. When differentiating a definite integral, we often apply what's known as the Leibniz Rule, an extension of the theorem for integrals with variable limits.

Here, when the upper limit of integration is a function of t, like \( \sqrt{t} \), we use it directly inside the integrand, multiplying by the derivative of the upper limit. This simplifies the process of differentiation considerably, avoiding lengthy calculations. Understanding this concept is crucial for moving seamlessly between integration and differentiation, as demonstrated by the steps in solution.
Definite Integral
A definite integral is a key concept representing the accumulation of quantities, such as area under a curve, over an interval \([a, b]\). Unlike an indefinite integral, it does not include a constant of integration because the result is a real number.

In solving the given exercise, we first evaluated a definite integral from 0 to \( \sqrt{t} \). This process provided the groundwork for finding a function that needed to be differentiated eventually. Here’s how it typically works:

  • Setup: Identify the limits and the function inside the integral.
  • Evaluate: Substitute the upper limit into the indefinite integral result, then subtract the result of substituting the lower limit.
This evaluation leads us to an expression involving \( t \) that we can later differentiate to find the derivative. This step is crucial because if we directly differentiate the integrand itself without considering the limits, we can make errors in results. By properly carrying out the definite integral, we prepare ourselves for an accurate application of the Fundamental Theorem of Calculus.
Chain Rule
The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. When a function is nested within another function, the Chain Rule provides a systematic method to find the derivative.

For example, in the differentiation of \( 3 \arcsin(\sqrt{t}) \), we utilize the Chain Rule because both the arcsine function and the square root function are involved. Generally, the Chain Rule is applied as follows:

  • Outer Derivative: Take the derivative of the outer function first.
  • Inner Derivative: Multiply by the derivative of the inner function.
In our scenario, we first differentiate \( 3 \arcsin(u) \) where \( u = \sqrt{t} \), resulting in \( 3 / \sqrt{1-u^2} \). Then we multiply by the derivative of \( \sqrt{t} \), which is \( 1/(2\sqrt{t}) \).

Thus, correctly using the Chain Rule ensures we accommodate eventual changes both inside and outside the main function, giving us the final derivative needed in our integral differentiation solution.

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Most popular questions from this chapter

Suppose that \(f\) has a positive derivative for all values of \(x\) and that \(f(1)=0 .\) Which of the following statements must be true of the function $$g(x)=\int_{0}^{x} f(t) d t ?$$. Give reasons for your answers. a. \(g\) is a differentiable function of \(x\). b. \(g\) is a continuous function of \(x\). c. The graph of \(g\) has a horizontal tangent at \(x=1\). d. \(g\) has a local maximum at \(x=1\). e. \(g\) has a local minimum at \(x=1\). f. The graph of \(g\) has an inflection point at \(x=1\). g. The graph of \(d g / d x\) crosses the \(x\) -axis at \(x=1\).

In Exercises \(119-122,\) you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$f(x)=x+\sin (2 x), \quad g(x)=x^{3}$$

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Suppose that a company's marginal revenue from the manufacture and sale of eggbeaters is $$\frac{d r}{d x}=2-2 /(x+1)^{2}$$ where \(r\) is measured in thousands of dollars and \(x\) in thousands of units. How much money should the company expect from a production run of \(x=3\) thousand eggbeaters? To find out, integrate the marginal revenue from \(x=0\) to \(x=3\).

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