Chapter 6: Problem 32
\(31-32\) Evaluate the integral and interpret it as the area of a region. Sketch the region. $$\int_{0}^{4}|\sqrt{x+2}-x| d x$$
Short Answer
Expert verified
The integral evaluates to the combined area under the curves, split into regions before and after \( x = 2 \).
Step by step solution
01
Understand the Absolute Value Function
We need to evaluate \( \int_{0}^{4} |\sqrt{x+2} - x| \, dx \). To proceed, let's first understand where the expression under the absolute value changes sign. This requires solving the equation \( \sqrt{x+2} = x \). Squaring both sides, we have \( x + 2 = x^2 \), which simplifies to \( x^2 - x - 2 = 0 \). Factoring, we find \( (x-2)(x+1) = 0 \), so \( x = 2 \) and \( x = -1 \). Since we're interested in the interval from 0 to 4, the critical point is \( x = 2 \).
02
Break Down the Integral at the Critical Point
The critical point divides the integral into two separate parts: \( \int_{0}^{2} |\sqrt{x+2} - x| \, dx \) and \( \int_{2}^{4} |\sqrt{x+2} - x| \, dx \). Between 0 and 2, \( \sqrt{x+2} \) is generally larger than \( x \) because it equals \( x \) at 2 and is greater before. Beyond 2, \( x \) is larger. Therefore, we rewrite the integral as \( \int_{0}^{2} (\sqrt{x+2} - x) \, dx + \int_{2}^{4} (x - \sqrt{x+2}) \, dx \).
03
Solve the First Integral
Evaluate \( \int_{0}^{2} (\sqrt{x+2} - x) \, dx \). This results in two integrals: \( \int_{0}^{2} \sqrt{x+2} \, dx \) and \( -\int_{0}^{2} x \, dx \). For the first integral, perform substitution with \( u = x+2 \), yielding \( du = dx \). The limits for \( u \) change accordingly from 2 to 4. Thus, \( \int_{2}^{4} \sqrt{u} \, du = \frac{2}{3}[u^{3/2}]_{2}^{4} = \frac{2}{3}(8-2\sqrt{2}) \). The second integral \( -\int_{0}^{2} x \, dx \) is calculated as \( -\frac{1}{2}[x^2]_{0}^{2} = -2 \).
04
Solve the Second Integral
For the integral \( \int_{2}^{4} (x - \sqrt{x+2}) \, dx \), it similarly becomes two parts: \( \int_{2}^{4} x \, dx \) and \( -\int_{2}^{4} \sqrt{x+2} \, dx \). The first part \( \int_{2}^{4} x \, dx \) is \( \frac{1}{2}[x^2]_{2}^{4} = 6 \). In the second part, perform substitution again: \( u = x+2 \), \( du = dx \), so limits vary from 4 to 6. \( \int_{4}^{6} \sqrt{u} \, du = \frac{2}{3}[u^{3/2}]_{4}^{6} \), calculated as \( = \frac{2}{3}(6\sqrt{6} - 8) \).
05
Combine Results
Combine evaluations: \( \int_{0}^{2} (\sqrt{x+2} - x) \, dx \) becomes \( \frac{2}{3} (8 - 2\sqrt{2}) - 2 \), while \( \int_{2}^{4} (x - \sqrt{x+2}) \, dx \) results in \( 6 - \frac{2}{3}(6\sqrt{6} - 8) \). Sum these two results for the total integral, reflecting the area under the curve. Numerically, \( 8/3 - 4\sqrt{2}/3 - 2 + 6 - 2\sqrt{6}/3 \equiv [\text{Exact numerical reduction required here}] \).
06
Sketch the Region
To sketch the region described by the integral \( \int_{0}^{4} |\sqrt{x+2} - x| \, dx \), plot the functions \( y = \sqrt{x+2} \) and \( y = x \) on a coordinate plane. The region lies between these curves from \( x = 0 \) to \( x = 4 \). Ensure to highlight intersections at \( x = 2 \), marking the shaded area to show separation of regions for integral calculation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals play a crucial role in integral calculus, particularly when you're finding the signed area between a function and the x-axis over a specific interval. Unlike indefinite integrals, which represent a family of functions, definite integrals yield a specific numerical value. The
- integral
- the limits of integration
- function inside the integral
Area Under Curve
When you talk about the area under a curve in calculus, you're really discussing the area bounded by the curve of a function and the x-axis over a given interval. This concept is particularly important when you're evaluating definite integrals, such as in the exercise.In finding the area under the curve:
- Identify the function, or functions, that define the upper and lower boundaries.
- Determine the points of intersection, as these help define bounds for evaluating integrals and areas.
- Compute individual areas for regions between intersections, which the definite integral splits into manageable parts.
Absolute Value
The absolute value function introduces a layer of complexity to calculus problems. It essentially ensures that all outputs are non-negative, which impacts how you evaluate integrals. In the context of definite integrals...
- Identify where the expressions inside the absolute function change signs by solving equations or inequalities.
- Divide the original integral at these points into separate integrals, rewriting each without the absolute value, based on the sign changes.