り6 C). 0
Angles in Circles
Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.
Arcs in Circles
A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.
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![### Calculus Problem: Evaluating an Integral
Consider the following integral:
\[ \int_{0}^{1} \frac{dx}{x^2 + 3x + 2} \]
Below are multiple choice options for the result of the integral:
a) \(\ln{\frac{2}{3}}\)
b) \(\infty\)
c) \(0\)
d) \(-\ln{\frac{3}{4}}\)
e) \(\ln{\frac{3}{4}}\)
### Explanation:
To evaluate the integral \(\int_{0}^{1} \frac{dx}{x^2 + 3x + 2}\), we can first factor the denominator:
\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]
Thus, the integral becomes:
\[ \int_{0}^{1} \frac{dx}{(x + 1)(x + 2)} \]
We can use partial fraction decomposition to simplify this further. We write:
\[ \frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \]
Solving for \(A\) and \(B\), we get:
\[ \frac{1}{(x + 1)(x + 2)} = \frac{A(x + 2) + B(x + 1)}{(x + 1)(x + 2)} \]
\[ 1 = A(x + 2) + B(x + 1) \]
By comparing coefficients or substituting suitable values of \(x\), we find \(A = 1\) and \(B = -1\). Therefore:
\[ \frac{1}{(x + 1)(x + 2)} = \frac{1}{x + 1} - \frac{1}{x + 2} \]
The integral can now be written as:
\[ \int_{0}^{1} \left(\frac{1}{x + 1} - \frac{1}{x + 2}\right) dx \]
This can be separated into two integrals:
\[ \int_{0}^{1} \frac{1}{x + 1} dx - \int_{0}^{1} \frac{1}{x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faaa612e4-bb6a-4a7c-9bcb-a4980796bdd1%2F728d5576-bdbc-472f-9f6b-887daaddef4f%2Fl7pfbe_reoriented.jpeg&w=3840&q=75)
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