Calculus 2 Final Exam Review

To find the area of the region bounded by the curve \( y = \frac{x}{\sqrt{x^2 + 5}} \), the x-axis, and the line \( x = 2 \), we need to set up an integral from \( x = 0 \) to \( x = 2 \). The area can be calculated as \( A = \int_0^2 \frac{x}{\sqrt{x^2 + 5}} \, dx \). Evaluating this integral yields the area, which approximates to 5.27 \( u^2 \). This result represents the area under the curve in the specified region.

To find the area between the curves \( y = x^3 \) and \( y = 4x \) in the first quadrant, we first determine their points of intersection by setting \( x^3 = 4x \). This leads to \( x(x^2 - 4) = 0 \), giving intersection points at \( x = 0 \) and \( x = 2 \). The area can be calculated using the integral \( \int_0^2 (4x - x^3) \, dx \). Evaluating this integral results in an area of 4 square units, representing the bounded region in the first quadrant.

To find the area of the region bounded by the curves \(x = y^2 - 4\) and \(y = x + 2\), we first determine their points of intersection. By substituting \(y\) from the second equation into the first, we can solve for \(y\) and subsequently for \(x\). The area between the curves can then be calculated using definite integrals, integrating the difference between the upper and lower functions over the interval defined by the intersection points. This results in an area of \(9/2\) square units.

To find the area between the curves \(y = \cos x\) and \(y = \sin x\) from \(x = 0\) to \(x = \frac{\pi}{4}\), we first determine the points of intersection, which occur at \(x = \frac{\pi}{4}\). The area can be calculated by integrating the difference of the two functions over the interval. Thus, the area \(A\) is given by:

\[
A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx
\]

Evaluating this integral yields \(A = \sqrt{2} - 1\) units², representing the area between the two curves in the specified interval.

To find the area between the parabola \( y = x^2 - 2x \) and the line \( y = x + 4 \), first determine their points of intersection by setting the equations equal to each other. Solving the resulting quadratic equation gives the bounds for integration. The area can then be calculated by integrating the difference of the line and the parabola over the interval defined by the intersection points. Evaluating this integral yields the area of the region, which is \( \frac{125}{6} \) square units.

To find the area between the curves \( y = \sqrt{x} \) and \( y = x^2 \), we first determine their points of intersection by setting \( \sqrt{x} = x^2 \). Solving this gives \( x = 0 \) and \( x = 1 \). The area between the curves from \( x = 0 \) to \( x = 1 \) is calculated using the integral of the top curve minus the bottom curve:

\[
\int_0^1 (\sqrt{x} - x^2) \, dx.
\]

Evaluating this integral yields \( \frac{1}{3} \) square units, representing the area of the region bounded by the two curves.

To find the area between the parabola \( y^2 = 2x - 2 \) and the line \( y = x - 5 \), first determine their points of intersection by substituting \( y \) from the line equation into the parabola's equation. This yields a quadratic equation in \( x \). Solving this gives the x-coordinates of the intersection points. The area can then be calculated using the definite integral of the upper curve (the line) minus the lower curve (the parabola) between these intersection points, resulting in an area of 18 square units.

To find the area between the curves \(y = x^2\) and \(y = -x^2 + 4x\), we first determine their points of intersection by setting the equations equal: \(x^2 = -x^2 + 4x\). Solving this gives the intersection points \(x = 0\) and \(x = 2\). The area can then be calculated using the integral of the upper curve minus the lower curve from 0 to 2:

\[
\int_0^2 ((-x^2 + 4x) - x^2) \,dx = \int_0^2 (-2x^2 + 4x) \,dx.
\]

Evaluating this integral yields \(8/3\) units², representing the area of the bounded region.

To find the area of the region bounded by the curve \( y = x^3 - 2x^2 - 5x + 6 \), the x-axis, and the vertical lines \( x = -1 \) and \( x = 2 \), we first determine the points where the curve intersects the x-axis by solving \( y = 0 \). Then, we compute the definite integral of the curve from \( x = -1 \) to \( x = 2 \). The resulting value gives the net area, which is positive because the curve is above the x-axis in this interval. The final area is \( \frac{157}{12} \) square units.

To find the area of the region bounded by the curve \( y = x^2 - 4x \), the x-axis, and the vertical lines \( x = 1 \) and \( x = 3 \), we first determine the points of intersection of the curve with the x-axis by setting \( y = 0 \). The area can be calculated using definite integration from \( x = 1 \) to \( x = 3 \). Evaluating the integral of the quadratic function over this interval gives the area as \( \frac{22}{3} \) square units.

To evaluate the integral ∫ √(9−x²)/x² dx, we can use integration techniques involving trigonometric substitutions or recognizing the form of the integral. The integrand can be transformed using the identity for arcsine and properties of square roots. The resulting expression simplifies to −√(9−x²)/x − arcsin(x/3) + C, where the first term represents the geometric interpretation of the integral and the second term arises from the integration of the arcsine function. This combination correctly captures the behavior of the integral over its domain.

To evaluate the integral ∫ (3x+7)/((x−3)(x+1)) dx, we can use partial fraction decomposition. This involves expressing the integrand as a sum of simpler fractions, which can then be integrated individually. After decomposing, integrating each term yields logarithmic functions. The specific coefficients in front of the logarithmic terms are determined by the decomposition process. The final result combines these logarithmic terms, leading to 4 ln|x−3| − ln|x+1| + C, where C is the constant of integration.

To evaluate the integral ∫ (2x+5)/(x+1)² dx, we can use the method of partial fractions or substitution. By rewriting the integrand, we can separate it into simpler components. The term (2x + 5) can be expressed in relation to the derivative of (x + 1)², leading to a logarithmic function. The integration results in 2 ln|x + 1| from the logarithmic part and a rational function component yielding -3/(x + 1). Combining these results gives the final answer, which includes the constant of integration C.

To evaluate the integral ∫ (5x+1)/((x−1)(x+2)) dx, we can use partial fraction decomposition. We express the integrand as A/(x-1) + B/(x+2) and solve for constants A and B. After determining these constants, we integrate each term separately. The integration yields logarithmic functions of the factors in the denominator. The final result combines these logarithmic terms, leading to the expression 2 ln|x−1| + 3 ln|x+2| + C, where C is the constant of integration.

To evaluate the integral ∫ 1/(x(x+3)) dx, we can use partial fraction decomposition to express the integrand as A/x + B/(x+3). Solving for A and B gives us the fractions needed for integration. After integrating each term, we find that the result involves the natural logarithm of the absolute value of the ratio of the factors, specifically (1/3) ln|x/(x+3)|. The constant C represents the integration constant, completing the solution.

To evaluate the integral ∫ 1/(x²−25) dx, we first recognize that the denominator can be factored as (x−5)(x+5). This suggests using partial fraction decomposition. We express the integrand as A/(x−5) + B/(x+5) and solve for A and B. After finding the constants, we integrate each term separately. The result involves the natural logarithm of the absolute values of the factors. The final expression simplifies to (1/10) ln|x−5|/|x+5| + C, reflecting the integration of the decomposed fractions.

To evaluate the integral ∫ √(x²−16)/x dx, we can use a trigonometric substitution, setting x = 4sec(θ), which simplifies the expression under the square root. After substituting and simplifying, integration yields a combination of the square root and the arcsecant function. The specific answer provided, √(x²−16) − 4 arcsec(x/4) + C, represents the result of this integration, where the arcsecant term accounts for the relationship between the variable x and the trigonometric function used in the substitution.

To evaluate the integral ∫₀² x³/√(16−x²) dx, we can use substitution and integration techniques. The expression under the square root suggests a trigonometric substitution, such as x = 4sin(θ), which simplifies the integral. After performing the substitution and integrating, we find the result involves evaluating definite limits. The final answer is expressed in terms of rational numbers and square roots, leading to the solution 128/3 − 24√3, which combines both components effectively.

To evaluate the integral ∫ dx / (x²√(4−x²)), we can use trigonometric substitution, letting x = 2sin(θ). This transforms the integral into a more manageable form. After simplifying, we find that the integral leads to the expression −√(4−x²)/(4x) + C. This result arises from the integration process, where we account for the substitution and the resulting changes in limits and differential. The negative sign reflects the orientation of the integral, while the constant C represents the arbitrary constant of integration.

To evaluate the integral ∫ x³/√(x²+9) dx, we can use integration techniques such as substitution or integration by parts. By substituting u = x² + 9, we simplify the expression and find that the integral leads to terms involving u^(3/2) and √(u). After performing the integration and simplifying, we arrive at the expression (1/3)(x²+9)^(3/2) − 9√(x²+9) + C, which includes both polynomial and radical components, reflecting the original integrand's structure.