Math / Calculus

QuestionEvaluate the following definite integral: To input the inverse triginometric functions $\cos ^{-1}$ and $\sin ^{-1}$ , type acos or asin respectively (short for arccos and arcsin). $\int_{2}^{5} \frac{\sqrt{x^{2}-4}}{x} d x=$ $\square$ Check

Studdy Solution

STEP 1

What is this asking? We need to find the area under the graph of $\frac{\sqrt{x^2 - 4}}{x}$ from $x = 2$ to $x = 5$ . Watch out! This integral involves a square root, so we'll likely use trigonometric substitution.
Don't forget to change the limits of integration when you substitute!

STEP 2

1. Trig Substitution
2. Simplify and Solve
3. Back-Substitute and Evaluate

STEP 3

Since we see $\sqrt{x^2 - 4}$ , a trigonometric substitution involving secant seems promising.
Let's try $x = 2\sec(\theta)$ .
This means $dx = 2\sec(\theta)\tan(\theta) d\theta$ .

STEP 4

When $x = 2$ , $2 = 2\sec(\theta)$ , so $\sec(\theta) = 1$ , meaning $\theta = 0$ .
When $x = 5$ , $5 = 2\sec(\theta)$ , so $\sec(\theta) = \frac{5}{2}$ , meaning $\theta = \sec^{-1}(\frac{5}{2})$ .
Let's call this value $\alpha$ for now, so $\alpha = \sec^{-1}(\frac{5}{2})$ .

STEP 5

Our integral becomes $\int_{0}^{\alpha} \frac{\sqrt{(2\sec(\theta))^2 - 4}}{2\sec(\theta)} \cdot 2\sec(\theta)\tan(\theta) d\theta$ $= \int_{0}^{\alpha} \sqrt{4\sec^2(\theta) - 4} \cdot \tan(\theta) d\theta$

STEP 6

We can simplify the square root: $\sqrt{4\sec^2(\theta) - 4} = \sqrt{4(\sec^2(\theta) - 1)} = \sqrt{4\tan^2(\theta)} = 2\tan(\theta)$ So our integral becomes: $\int_{0}^{\alpha} 2\tan(\theta) \cdot \tan(\theta) d\theta = 2\int_{0}^{\alpha} \tan^2(\theta) d\theta$

STEP 7

We know that $\tan^2(\theta) = \sec^2(\theta) - 1$ , so our integral becomes: $2\int_{0}^{\alpha} (\sec^2(\theta) - 1) d\theta = 2[\tan(\theta) - \theta]_{0}^{\alpha}$

STEP 8

We have $2[\tan(\alpha) - \alpha] - 2[\tan(0) - 0] = 2(\tan(\alpha) - \alpha)$ , since $\tan(0) = 0$ .

STEP 9

Remember, $\sec(\alpha) = \frac{5}{2}$ .
If we draw a right triangle with angle $\alpha$ , the adjacent side is 2 and the hypotenuse is 5.
Using the Pythagorean theorem, the opposite side is $\sqrt{5^2 - 2^2} = \sqrt{21}$ .
So, $\tan(\alpha) = \frac{\sqrt{21}}{2}$ .
Also, $\alpha = \sec^{-1}(\frac{5}{2})$ .

STEP 10

Our final answer is $2(\frac{\sqrt{21}}{2} - \sec^{-1}(\frac{5}{2})) = \sqrt{\textbf{21}} - 2\sec^{-1}(\frac{\textbf{5}}{\textbf{2}})$ .

STEP 11

The value of the definite integral is $\sqrt{21} - 2\sec^{-1}(\frac{5}{2})$ .

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QuestionEvaluate the following definite integral: To input the inverse triginometric functions $\cos ^{-1}$ and $\sin ^{-1}$ , type acos or asin respectively (short for arccos and arcsin). $\int_{2}^{5} \frac{\sqrt{x^{2}-4}}{x} d x=$ $\square$ Check

Studdy Solution

STEP 1

What is this asking? We need to find the area under the graph of $\frac{\sqrt{x^2 - 4}}{x}$ from $x = 2$ to $x = 5$ . Watch out! This integral involves a square root, so we'll likely use trigonometric substitution.
Don't forget to change the limits of integration when you substitute!

STEP 2

1. Trig Substitution
2. Simplify and Solve
3. Back-Substitute and Evaluate

STEP 3

Since we see $\sqrt{x^2 - 4}$ , a trigonometric substitution involving secant seems promising.
Let's try $x = 2\sec(\theta)$ .
This means $dx = 2\sec(\theta)\tan(\theta) d\theta$ .

STEP 4

When $x = 2$ , $2 = 2\sec(\theta)$ , so $\sec(\theta) = 1$ , meaning $\theta = 0$ .
When $x = 5$ , $5 = 2\sec(\theta)$ , so $\sec(\theta) = \frac{5}{2}$ , meaning $\theta = \sec^{-1}(\frac{5}{2})$ .
Let's call this value $\alpha$ for now, so $\alpha = \sec^{-1}(\frac{5}{2})$ .

STEP 5

Our integral becomes $\int_{0}^{\alpha} \frac{\sqrt{(2\sec(\theta))^2 - 4}}{2\sec(\theta)} \cdot 2\sec(\theta)\tan(\theta) d\theta$ $= \int_{0}^{\alpha} \sqrt{4\sec^2(\theta) - 4} \cdot \tan(\theta) d\theta$

STEP 6

We can simplify the square root: $\sqrt{4\sec^2(\theta) - 4} = \sqrt{4(\sec^2(\theta) - 1)} = \sqrt{4\tan^2(\theta)} = 2\tan(\theta)$ So our integral becomes: $\int_{0}^{\alpha} 2\tan(\theta) \cdot \tan(\theta) d\theta = 2\int_{0}^{\alpha} \tan^2(\theta) d\theta$

STEP 7

We know that $\tan^2(\theta) = \sec^2(\theta) - 1$ , so our integral becomes: $2\int_{0}^{\alpha} (\sec^2(\theta) - 1) d\theta = 2[\tan(\theta) - \theta]_{0}^{\alpha}$

STEP 8

We have $2[\tan(\alpha) - \alpha] - 2[\tan(0) - 0] = 2(\tan(\alpha) - \alpha)$ , since $\tan(0) = 0$ .

STEP 9

STEP 10

Our final answer is $2(\frac{\sqrt{21}}{2} - \sec^{-1}(\frac{5}{2})) = \sqrt{\textbf{21}} - 2\sec^{-1}(\frac{\textbf{5}}{\textbf{2}})$ .

STEP 11

The value of the definite integral is $\sqrt{21} - 2\sec^{-1}(\frac{5}{2})$ .

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Studdy Solution

STEP 1

STEP 2

1. Trig Substitution2. Simplify and Solve3. Back-Substitute and Evaluate

STEP 3

Since we see x2−4\sqrt{x^2 - 4}x2−4​, a trigonometric substitution involving secant seems promising.Let's try x=2sec⁡(θ)x = 2\sec(\theta)x=2sec(θ).This means dx=2sec⁡(θ)tan⁡(θ)dθdx = 2\sec(\theta)\tan(\theta) d\thetadx=2sec(θ)tan(θ)dθ.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

We have 2[tan⁡(α)−α]−2[tan⁡(0)−0]=2(tan⁡(α)−α)2[\tan(\alpha) - \alpha] - 2[\tan(0) - 0] = 2(\tan(\alpha) - \alpha)2[tan(α)−α]−2[tan(0)−0]=2(tan(α)−α), since tan⁡(0)=0\tan(0) = 0tan(0)=0.

STEP 9

STEP 10

Our final answer is 2(212−sec⁡−1(52))=21−2sec⁡−1(52)2(\frac{\sqrt{21}}{2} - \sec^{-1}(\frac{5}{2})) = \sqrt{\textbf{21}} - 2\sec^{-1}(\frac{\textbf{5}}{\textbf{2}})2(221​​−sec−1(25​))=21​−2sec−1(25​).

STEP 11

The value of the definite integral is 21−2sec⁡−1(52)\sqrt{21} - 2\sec^{-1}(\frac{5}{2})21​−2sec−1(25​).

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1. Trig Substitution
2. Simplify and Solve
3. Back-Substitute and Evaluate

Since we see $\sqrt{x^2 - 4}$ , a trigonometric substitution involving secant seems promising.
Let's try $x = 2\sec(\theta)$ .
This means $dx = 2\sec(\theta)\tan(\theta) d\theta$ .

We have $2[\tan(\alpha) - \alpha] - 2[\tan(0) - 0] = 2(\tan(\alpha) - \alpha)$ , since $\tan(0) = 0$ .

Our final answer is $2(\frac{\sqrt{21}}{2} - \sec^{-1}(\frac{5}{2})) = \sqrt{\textbf{21}} - 2\sec^{-1}(\frac{\textbf{5}}{\textbf{2}})$ .

The value of the definite integral is $\sqrt{21} - 2\sec^{-1}(\frac{5}{2})$ .