Evaluate The Integral Of Sqrt(5 - 4x - X^2) From -2 To 1

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#seo title: Evaluating the Integral of sqrt(5 - 4x - x^2) from -2 to 1

In this comprehensive guide, we will meticulously evaluate the definite integral ${ \int_{-2}^{1} \sqrt{5 - 4x - x^2} \, dx }$. This integral might appear daunting at first glance, but by employing strategic techniques such as completing the square and trigonometric substitution, we can systematically break it down and arrive at the solution. Our exploration will not only provide the answer but also illuminate the underlying principles and methodologies applicable to a broader range of integral problems.

1. Transforming the Integrand: Completing the Square

The initial step in tackling this integral involves transforming the integrand, specifically the expression under the square root, into a more manageable form. The expression ${ 5 - 4x - x^2 }$ is a quadratic, and completing the square is the ideal technique for rewriting it. This process allows us to express the quadratic as a squared term plus a constant, which is far more amenable to trigonometric substitution. By identifying the quadratic expression under the square root, we aim to rewrite it in a form that facilitates integration. This involves algebraic manipulation to reveal a structure that fits a known trigonometric identity. We start by rearranging the terms and factoring out a negative sign:

${ 5 - 4x - x^2 = -(x^2 + 4x) + 5 }$

Now, we focus on the expression inside the parentheses, ${ x^2 + 4x }$. To complete the square, we take half of the coefficient of the ${ x }$ term (which is 4), square it (resulting in 4), and add it inside the parentheses. Of course, to maintain the equality, we must also subtract the same amount (multiplied by the negative sign we factored out) outside the parentheses:

${ -(x^2 + 4x) + 5 = -(x^2 + 4x + 4) + 5 + 4 }$

This simplifies to:

${ -(x + 2)^2 + 9 }$

Thus, the expression under the square root becomes ${ 9 - (x + 2)^2 }$. This transformation is crucial because it now closely resembles the form ${ a^2 - u^2 }$, which is associated with the trigonometric identity ${ ext{sin}^2( heta) + ext{cos}^2( heta) = 1 }$. Completing the square not only simplifies the algebraic structure but also paves the way for a trigonometric substitution that will untangle the integral. This meticulous algebraic manipulation is a powerful tool in calculus, enabling us to approach complex integrands with a clear strategy.

2. Strategic Trigonometric Substitution: A Gateway to Simplification

Having successfully completed the square, our integrand now features the expression ${ \sqrt{9 - (x + 2)^2} }$. This form strongly suggests a trigonometric substitution to further simplify the integral. Trigonometric substitution is a powerful technique that leverages trigonometric identities to transform complex integrals into more manageable forms. The key is to recognize the structural similarity between the integrand and standard trigonometric expressions. In this case, the expression resembles the form ${ \sqrt{a^2 - u^2} }$, where ${ a = 3 }$ and ${ u = x + 2 }$. This pattern directly corresponds to the Pythagorean identity involving sine and cosine.

Specifically, we will employ the substitution:

${ x + 2 = 3 ext{sin}(\theta) }$

This substitution is motivated by the identity ${ 1 - ext{sin}^2(\theta) = ext{cos}^2(\theta) }$. By making this substitution, we aim to eliminate the square root and express the integrand in terms of trigonometric functions. The next step is to find the differential ${ dx }$ in terms of ${ d\theta }$. Differentiating both sides of the substitution equation with respect to ${ \theta }$ yields:

${ dx = 3 ext{cos}(\theta) \, d\theta }$

Furthermore, we need to change the limits of integration to correspond to the new variable ${ \theta }$. When ${ x = -2 }$, we have:

${ -2 + 2 = 3 ext{sin}(\theta) \implies ext{sin}(\theta) = 0 \implies \theta = 0 }$

And when ${ x = 1 }$, we have:

${ 1 + 2 = 3 ext{sin}(\theta) \implies ext{sin}(\theta) = 1 \implies \theta = \frac{\pi}{2} }$

Thus, our new limits of integration are 0 and ${ \frac{\pi}{2} }$. Now, we can substitute everything back into the integral, replacing ${ x + 2 }$ with ${ 3 ext{sin}(\theta) }$, ${ dx }$ with ${ 3 ext{cos}(\theta) \, d\theta }$, and adjusting the limits of integration accordingly. This strategic substitution is a crucial step in transforming the integral into a form that we can readily evaluate.

3. Integral Transformation: Unveiling the Trigonometric Form

With the trigonometric substitution in place, we can now transform the original integral into a trigonometric form that is much easier to handle. Recall that we have substituted ${ x + 2 = 3 ext{sin}(\theta) }$ and ${ dx = 3 ext{cos}(\theta) \, d\theta }$. Our original integral was:

${ \int_{-2}^{1} \sqrt{5 - 4x - x^2} \, dx }$

After completing the square and applying the substitution, the integral becomes:

${ \int_{0}^{\frac{\pi}{2}} \sqrt{9 - (3 ext{sin}(\theta))^2} \cdot 3 ext{cos}(\theta) \, d\theta }$

First, let's simplify the expression under the square root:

${ \sqrt{9 - 9 ext{sin}^2(\theta)} = \sqrt{9(1 - ext{sin}^2(\theta))} = 3\sqrt{\text{cos}^2(\theta)} = 3 ext{cos}(\theta) }$

Assuming ${ 0 \leq \theta \leq \frac{\pi}{2} }$, ${ ext{cos}(\theta) }$ is non-negative, so we can take the positive square root. Now, substituting this back into the integral, we get:

${ \int_{0}^{\frac{\pi}{2}} 3 ext{cos}(\theta) \cdot 3 ext{cos}(\theta) \, d\theta = 9 \int_{0}^{\frac{\pi}{2}} ext{cos}^2(\theta) \, d\theta }$

This transformed integral is significantly simpler than the original. The square root has been eliminated, and we are left with the integral of ${ ext{cos}^2(\theta) }$, a standard trigonometric integral. This transformation highlights the power of trigonometric substitution in converting complex algebraic integrals into more manageable trigonometric forms. The next step will involve evaluating this trigonometric integral, which will ultimately lead us to the solution of the original problem.

4. Evaluating the Trigonometric Integral: A Path to the Solution

We have now arrived at the trigonometric integral ${ 9 \int_{0}^{\frac{\pi}{2}} ext{cos}^2(\theta) \, d\theta }$. To evaluate this, we will employ a trigonometric identity to rewrite ${ ext{cos}^2(\theta) }$. The identity we need is the double-angle formula for cosine:

${ ext{cos}(2\theta) = 2 ext{cos}^2(\theta) - 1 }$

Solving for ${ ext{cos}^2(\theta) }$, we get:

${ ext{cos}^2(\theta) = \frac{1 + ext{cos}(2\theta)}{2} }$

Substituting this into our integral, we have:

${ 9 \int_{0}^{\frac{\pi}{2}} \frac{1 + ext{cos}(2\theta)}{2} \, d\theta }$

We can now split this integral into two parts:

${ \frac{9}{2} \int_{0}^{\frac{\pi}{2}} (1 + ext{cos}(2\theta)) \, d\theta = \frac{9}{2} \left[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta + \int_{0}^{\frac{\pi}{2}} ext{cos}(2\theta) \, d\theta \right] }$

The first integral is straightforward:

${ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = [\theta]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} }$

For the second integral, we use a simple u-substitution. Let ${ u = 2\theta }$, so ${ du = 2 \, d\theta }$ and ${ d\theta = \frac{1}{2} \, du }$. The new limits of integration are ${ u = 2(0) = 0 }$ and ${ u = 2(\frac{\pi}{2}) = \pi }$. Thus,

${ \int_{0}^{\frac{\pi}{2}} ext{cos}(2\theta) \, d\theta = \frac{1}{2} \int_{0}^{\pi} ext{cos}(u) \, du = \frac{1}{2} [\text{sin}(u)]_{0}^{\pi} = \frac{1}{2} (\text{sin}(\pi) - \text{sin}(0)) = 0 }$

Now, we can substitute these results back into the original expression:

${ \frac{9}{2} \left[ \frac{\pi}{2} + 0 \right] = \frac{9\pi}{4} }$

Thus, the value of the integral is ${ \frac{9\pi}{4} }$. This step-by-step evaluation demonstrates how trigonometric identities and u-substitution can be used to solve trigonometric integrals, leading us to the final numerical answer.

5. The Final Verdict: Unveiling the Definite Integral's Value

After meticulously navigating through the steps of completing the square, applying trigonometric substitution, and evaluating the resulting trigonometric integral, we have arrived at the solution. The definite integral ${ \int_{-2}^{1} \sqrt{5 - 4x - x^2} \, dx }$ has been shown to be equal to ${ \frac{9\pi}{4} }$. This result represents the exact value of the integral, a testament to the power and precision of calculus techniques.

Let's recap the key stages of our journey. First, we transformed the integrand by completing the square, converting the expression under the square root into a form suitable for trigonometric substitution. This algebraic manipulation was crucial in simplifying the integral. Next, we employed a trigonometric substitution, replacing ${ x + 2 }$ with ${ 3 ext{sin}(\theta) }$. This substitution allowed us to eliminate the square root and express the integral in terms of trigonometric functions. We then transformed the integral, carefully changing the limits of integration to correspond to the new variable ${ \theta }$. The resulting trigonometric integral was significantly easier to handle. Finally, we evaluated the trigonometric integral using a double-angle identity and a simple u-substitution. This led us to the numerical answer of ${ \frac{9\pi}{4} }$.

The value ${ \frac{9\pi}{4} }$ holds a geometric interpretation as well. The original integral represents the area under the curve ${ y = \sqrt{5 - 4x - x^2} }$ between ${ x = -2 }$ and ${ x = 1 }$. The equation ${ y = \sqrt{5 - 4x - x^2} }$ describes the upper half of a circle centered at ${ (-2, 0) }$ with a radius of 3. The limits of integration, ${ -2 }$ and ${ 1 }$, correspond to the leftmost point and a point three-fourths of the way around the circle. Therefore, the integral calculates the area of three-quarters of a circle with radius 3, which is indeed ${ \frac{3}{4} \pi (3^2) = \frac{27\pi}{4} }$. The portion between -2 and 1 corresponds to half circle plus a quarter circle whose area is 3/4 of the total circle area, and the area of total circle is ${ \pi r^2 }$ with ${ r=3 }$, so ${ \pi 3^2 = 9 \pi }$, and 3/4 of it is ${ \frac{27\pi}{4} }$, so there might be a mistake in the previous calculations. Let's find the mistake in previous calculations. ${ y = \sqrt{5 - 4x - x^2} }$ can be written as ${ y^2 = 5 - 4x - x^2 }$, then ${ x^2 + 4x + y^2 = 5 }$. Completing the square gives ${ x^2 + 4x + 4 + y^2 = 5 + 4 }$, so ${ (x+2)^2 + y^2 = 9 = 3^2 }$. This is a circle centered at (-2, 0) with a radius of 3. Since y is the positive square root, we're considering the upper semicircle. The limits of integration are from x = -2 to x = 1. When x = -2, ${ (x+2)^2 = 0 }$, so ${ y^2 = 9 }$ and ${ y = 3 }$. When x = 1, ${ (1+2)^2 + y^2 = 9 }$, so ${ 9 + y^2 = 9 }$, then ${ y = 0 }$. The integral calculates the area of the upper semicircle from x = -2 to x = 1. This is 3/4 of the circle, since ${ sin(\theta) = (x+2)/3 }$, ${ x = -2 \implies sin(\theta) = 0 \implies \theta = 0 }$, ${ x = 1 \implies sin(\theta) = 1 \implies \theta = \pi /2 }$, So the angle varies from 0 to ${ \pi/2 }$, which is 1/4 of full circle, so we take 3 times of the area in 1/4 circle which make it 3/4 full circle. so area should be 3/4 * ${ \pi * 3^2 }$ = ${ 27 \pi / 4 }$.

In conclusion, evaluating the definite integral ${ \int_{-2}^{1} \sqrt{5 - 4x - x^2} \, dx }$ exemplifies the elegance and power of calculus. By systematically applying techniques such as completing the square and trigonometric substitution, we can solve complex integrals and gain a deeper appreciation for the interconnectedness of algebra, trigonometry, and calculus.