Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Certainly! Let's solve the integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex].
First, identify the integrand: [tex]\(\frac{2x + 1}{\sqrt{x^2 + 4}}\)[/tex].
To solve this integral, we can use techniques such as substitution and recognizing standard forms of integrals. Here is a detailed step-by-step solution:
1. Recognize that the expression under the square root [tex]\(x^2 + 4\)[/tex] suggests using a trigonometric or hyperbolic substitution. However, we can also consider a more straightforward method by splitting the integrand.
2. Split the integrand:
[tex]\[ \int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx = \int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx + \int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx \][/tex]
3. Solve the first part [tex]\(\int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(u = x^2 + 4\)[/tex]. Then, [tex]\(du = 2x \, dx\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(u = 4\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(u = 8\)[/tex].
The integral now becomes:
[tex]\[ \int_{4}^{8} \frac{1}{\sqrt{u}} \, du = \int_{4}^{8} u^{-\frac{1}{2}} \, du \][/tex]
Evaluate the integral:
[tex]\[ \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} \][/tex]
So we have:
[tex]\[ 2u^{\frac{1}{2}} \Big|_4^8 = 2\left(8^{\frac{1}{2}} - 4^{\frac{1}{2}}\right) = 2(\sqrt{8} - \sqrt{4}) = 2(2\sqrt{2} - 2) = 4\sqrt{2} - 4 \][/tex]
4. Solve the second part [tex]\(\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(x = 2 \tan \theta\)[/tex]. Then, [tex]\(dx = 2 \sec^2 \theta \, d\theta\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(\theta = 0\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(\theta = \frac{\pi}{4}\)[/tex].
The integral now becomes:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{4 \tan^2 \theta + 4}} \cdot 2 \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta}{2 \sec \theta} \, d\theta = \int_{0}^{\frac{\pi}{4}} \sec \theta \, d\theta \][/tex]
Evaluate the integral:
[tex]\[ \int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| \][/tex]
So we have:
[tex]\[ \ln|\sec \theta + \tan \theta| \Big|_0^{\frac{\pi}{4}} = \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)| = \ln|\sqrt{2} + 1| - \ln|1| = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1) \][/tex]
Combining both parts, the total value of the integral is:
[tex]\[ 4\sqrt{2} - 4 + \ln(\sqrt{2} + 1) \approx 2.53822783651192 \][/tex]
So, the evaluated integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex] equals [tex]\(\boxed{2.53822783651192}\)[/tex].
First, identify the integrand: [tex]\(\frac{2x + 1}{\sqrt{x^2 + 4}}\)[/tex].
To solve this integral, we can use techniques such as substitution and recognizing standard forms of integrals. Here is a detailed step-by-step solution:
1. Recognize that the expression under the square root [tex]\(x^2 + 4\)[/tex] suggests using a trigonometric or hyperbolic substitution. However, we can also consider a more straightforward method by splitting the integrand.
2. Split the integrand:
[tex]\[ \int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx = \int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx + \int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx \][/tex]
3. Solve the first part [tex]\(\int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(u = x^2 + 4\)[/tex]. Then, [tex]\(du = 2x \, dx\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(u = 4\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(u = 8\)[/tex].
The integral now becomes:
[tex]\[ \int_{4}^{8} \frac{1}{\sqrt{u}} \, du = \int_{4}^{8} u^{-\frac{1}{2}} \, du \][/tex]
Evaluate the integral:
[tex]\[ \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} \][/tex]
So we have:
[tex]\[ 2u^{\frac{1}{2}} \Big|_4^8 = 2\left(8^{\frac{1}{2}} - 4^{\frac{1}{2}}\right) = 2(\sqrt{8} - \sqrt{4}) = 2(2\sqrt{2} - 2) = 4\sqrt{2} - 4 \][/tex]
4. Solve the second part [tex]\(\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(x = 2 \tan \theta\)[/tex]. Then, [tex]\(dx = 2 \sec^2 \theta \, d\theta\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(\theta = 0\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(\theta = \frac{\pi}{4}\)[/tex].
The integral now becomes:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{4 \tan^2 \theta + 4}} \cdot 2 \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta}{2 \sec \theta} \, d\theta = \int_{0}^{\frac{\pi}{4}} \sec \theta \, d\theta \][/tex]
Evaluate the integral:
[tex]\[ \int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| \][/tex]
So we have:
[tex]\[ \ln|\sec \theta + \tan \theta| \Big|_0^{\frac{\pi}{4}} = \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)| = \ln|\sqrt{2} + 1| - \ln|1| = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1) \][/tex]
Combining both parts, the total value of the integral is:
[tex]\[ 4\sqrt{2} - 4 + \ln(\sqrt{2} + 1) \approx 2.53822783651192 \][/tex]
So, the evaluated integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex] equals [tex]\(\boxed{2.53822783651192}\)[/tex].
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
