Sketch the regions represented by the following integrals and evaluate them using geometric formulas.
4 integral 1: 2x+3dx
6 integral -2: lx-1l dx (absolute value of x-1
do you mean
4* integral dx/(2x+3)
or what?
4∫1 4 on the top and 1 on the bottom of that sign
6∫-2 same deal
Just use the power rule for these.
∫[1,4] 2x+3 dx = x^2+3x = 24
Recall the definition of |n|:
= n if n >= 0
= -n if n < 0
∫[-2,6] |x-1| dx has to be broken up at x=1, so you get
∫[-2,1] -(x-1) dx + ∫[1,6] (x+1) dx = 17
Thank you so much.
What would the sketched regions look like??
You're in calculus and can't sketch linear functions?
If you type in the integrals as I showed them, wolframalpha.com understands the notation. For example
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B-2,6%5D+%7Cx-1%7C+dx
To sketch the regions represented by the given integrals, we need to understand the integrands and the intervals of integration.
For integral 1: ∫(2x + 3) dx
The integrand is 2x + 3, which represents a linear function. To sketch the region, start by identifying the y-intercept, which is 3 in this case. Then, determine the slope, which is 2. This means that for every unit increase in x, the y-value will increase by 2. With this information, you can draw a straight line on the coordinate plane. To find the area under this curve, you need to know the interval of integration.
For integral 2: ∫[−2] (|x - 1|) dx
The integrand is |x - 1|, which represents the absolute value of x - 1. The absolute value function takes the value inside the absolute value bars and makes it positive. So, we have two cases to consider:
Case 1: x - 1 ≥ 0 or x ≥ 1
When x is greater than or equal to 1, the absolute value of x - 1 will be x - 1 itself. Therefore, the integrand becomes x - 1 in this region.
Case 2: x - 1 < 0 or x < 1
When x is less than 1, the absolute value of x - 1 will be -(x - 1), which becomes -x + 1. Thus, the integrand becomes -x + 1 in this region.
Now, let's sketch the two cases separately:
Case 1: x ≥ 1
For x ≥ 1, the integrand is x - 1. Graphically, this represents a straight line with a slope of 1 and a y-intercept of -1. We start the graph from the point (1, 0), since at x = 1 the value of |x - 1| becomes 0.
Case 2: x < 1
For x < 1, the integrand is -x + 1. This equation represents a straight line with a slope of -1 and a y-intercept of 1. We start the graph from the point (1, 0) as well, with the understanding that the line will be below the x-axis.
Now, to evaluate the integrals using geometric formulas:
For integral 1: ∫(2x + 3) dx
Using the geometric formula for the integral of a linear function f(x) = mx + b over an interval [a, b], we have:
∫(2x + 3) dx = [(1/2)(2x^2) + 3x] evaluated from a to b
= [(x^2) + 3x] evaluated from a to b
To evaluate the integral, substitute the limits of integration into the expression and subtract the evaluation of the lower limit from the evaluation of the upper limit.
For integral 2: ∫[−2] (|x - 1|) dx
Since we have divided the integral into two cases, we need to evaluate the integral separately for each case and then sum the results.
∫[−2] (|x - 1|) dx = ∫[-2 to 1] (x - 1) dx + ∫[1 to -2] (-x + 1) dx
= [(1/2)(x^2 - 2x)] evaluated from -2 to 1 + [(x - 1)] evaluated from 1 to -2
Now, you can substitute the limits of integration into the expressions and calculate the values to obtain the final results for both integrals.