The result of the definite integral is given by:
[tex]\frac{7\pi}{6}[/tex]
What is the solution to the definite integral?
The integral is given by:
[tex]\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{7}{1 + x^2} dx = 7 \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1 + x^2}[/tex]
Looking at an integral table, we have that:
[tex]\int \frac{dx}{1 + x^2} = \arctan{x}[/tex]
Hence the result of our integral is:
[tex]7 \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1 + x^2} = 7\arctan{x}|_{x = \frac{1}{\sqrt{3}}}^{x = \sqrt{3}}[/tex]
Considering the inverse trigonometric functions, we have that:
- [tex]\arctan{\frac{1}{\sqrt{3}}} = \frac{\pi}{6}[/tex], as [tex]\tan{\left(\frac{\pi}{6}\right)} = \frac{1}{\sqrt{3}}[/tex]
- [tex]\arctan{\sqrt{3}} = \frac{\pi}{3}[/tex], as [tex]\tan{\left(\frac{\pi}{3}\right)} = \sqrt{3}[/tex]
Hence, applying the Fundamental Theorem of Calculus, the result of the integral is:
[tex]7\arctan{x}|_{x = \frac{1}{\sqrt{3}}}^{x = \sqrt{3}} = 7\arctan{\sqrt{3}} - 7\arctan{\frac{1}{\sqrt{3}}} = 7\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{7\pi}{6}[/tex]
More can be learned about definite integrals at https://brainly.com/question/25706129
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