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Chapter 7: Problem 12

Evaluate the following integrals. $$\int_{-5}^{0} \frac{d x}{\sqrt{4-x}}$$

Short Answer

Expert verified
Question: Calculate the definite integral $\int_{-5}^{0} \frac{d x}{\sqrt{4-x}}$ Answer: The value of the definite integral is $\frac{3\pi}{2}$.

Step by step solution

01

Simplify the function in the integral

Since the denominator involves a square root, it's helpful to rewrite the expression in a more convenient form. We can do this by factoring out a negative sign in the square root: $$\frac{1}{\sqrt{4-x}} = \frac{1}{\sqrt{-(x-4)}}$$
02

Identify the antiderivative

With the new form of the function, we can now identify the antiderivative more easily. As the square root is in the denominator, we recognize this as the derivative of the arcsine function. Hence, the antiderivative of our function is: $$\int \frac{1}{\sqrt{-(x-4)}} dx = 2\arcsin\left(\sqrt{\frac{x-4}{4}}\right)$$
03

Evaluate the definite integral

Now we will apply the fundamental theorem of calculus by taking the antiderivative of the function. $$\int_{-5}^{0} \frac{dx}{\sqrt{4-x}} = \left[2\arcsin\left(\sqrt{\frac{x-4}{4}}\right)\right]_{-5}^{0}$$
04

Calculate the value at the limits

First, we'll calculate the value at x = 0: $$2\arcsin\left(\sqrt{\frac{0-4}{4}}\right) = 2\arcsin\left(\sqrt{-1}\right) = 2\arcsin\left(1\right) = 2\left(\frac{\pi}{2}\right) = \pi$$ Next, we'll calculate the value at x = -5: $$2\arcsin\left(\sqrt{\frac{-5-4}{4}}\right) = 2\arcsin\left(\sqrt{\frac{-1}{2}}\right) = 2\arcsin\left(-\frac{1}{\sqrt{2}}\right) = 2\left(-\frac{\pi}{4}\right) = -\frac{\pi}{2}$$
05

Subtract the values

Finally, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: $$\pi-(-\frac{\pi}{2}) = \pi+\frac{\pi}{2} = \frac{3\pi}{2}$$ The value of the definite integral is: $$\int_{-5}^{0} \frac{d x}{\sqrt{4-x}} = \frac{3\pi}{2}$$

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Most popular questions from this chapter

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{|x|} d x \quad\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-a x^{2}}.\) a. Graph the Gaussian for \(a=0.5,1,\) and 2. b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.

Use the Trapezoid Rule (Section 7 ) to approximate \(\int_{0}^{R} e^{-x^{2}} d x\) with \(R=2,4,\) and 8. For each value of \(R\), take \(n=4,8,16,\) and \(32,\) and compare approximations with successive values of \(n .\) Use these approximations to approximate \(I=\int_{0}^{\infty} e^{-x^{2}} d x.\)

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty).\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)
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