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Math  /  Calculus

Question16. Find possible values of aa and bb that make the statement true. If possible, use a sketch to support your answer. (There may be more than one correct answer.) a. ∫−21f(x)dx+∫15f(x)dx=∫abf(x)dx\int_{-2}^{1} f(x) d x+\int_{1}^{5} f(x) d x=\int_{a}^{b} f(x) d x b. ∫−33f(x)dx+∫36f(x)dx−∫abf(x)dx=∫−16f(x)dx\int_{-3}^{3} f(x) d x+\int_{3}^{6} f(x) d x-\int_{a}^{b} f(x) d x=\int_{-1}^{6} f(x) d x c. ∫absin⁡xdx<0\int_{a}^{b} \sin x d x<0 d. ∫abcos⁡xdx=0\int_{a}^{b} \cos x d x=0
17. The graph of ff consists of line segments, as show in the figure below. Evaluate each definite integral by using geometric formulas. A. ∫01−f(x)dx\int_{0}^{1}-f(x) d x B. ∫110f(x)dx\int_{11}^{0} f(x) d x C. ∫346−f(x)dx\int_{3}^{4} 6-f(x) d x D. ∫1043f(x)dx\int_{10}^{4} 3 f(x) d x

Studdy Solution

STEP 1

What is this asking? We need to find values for aa and bb that make some integral statements true, and then evaluate some definite integrals based on a graph of a function. Watch out! Remember the properties of integrals, like how the integral of a sum is the sum of the integrals, and how swapping the bounds of integration introduces a negative sign.
Also, don't forget that definite integrals represent areas under a curve!

STEP 2

1. Solve for *a* and *b* in the integral equations.
2. Evaluate the definite integrals using the graph.

STEP 3

We're given ∫−21f(x)dx+∫15f(x)dx=∫abf(x)dx\int_{-2}^{1} f(x) dx + \int_{1}^{5} f(x) dx = \int_{a}^{b} f(x) dx.
One of the coolest properties of integrals is that we can combine them!
If the upper bound of one integral matches the lower bound of the next, we can combine them into a single integral.
So, ∫−21f(x)dx+∫15f(x)dx\int_{-2}^{1} f(x) dx + \int_{1}^{5} f(x) dx becomes ∫−25f(x)dx\int_{-2}^{5} f(x) dx.
This means ∫−25f(x)dx=∫abf(x)dx\int_{-2}^{5} f(x) dx = \int_{a}^{b} f(x) dx.
Therefore, a=−2a = \mathbf{-2} and b=5b = \mathbf{5}.

STEP 4

We have ∫−33f(x)dx+∫36f(x)dx−∫abf(x)dx=∫−16f(x)dx\int_{-3}^{3} f(x) dx + \int_{3}^{6} f(x) dx - \int_{a}^{b} f(x) dx = \int_{-1}^{6} f(x) dx.
Let's combine the first two integrals: ∫−33f(x)dx+∫36f(x)dx=∫−36f(x)dx\int_{-3}^{3} f(x) dx + \int_{3}^{6} f(x) dx = \int_{-3}^{6} f(x) dx.
Now we have ∫−36f(x)dx−∫abf(x)dx=∫−16f(x)dx\int_{-3}^{6} f(x) dx - \int_{a}^{b} f(x) dx = \int_{-1}^{6} f(x) dx.
If we let a=−3a = \mathbf{-3} and b=−1b = \mathbf{-1}, then ∫−3−1f(x)dx=∫−36f(x)dx+∫6−1f(x)dx\int_{-3}^{-1} f(x) dx = \int_{-3}^{6} f(x) dx + \int_{6}^{-1} f(x) dx.
Since ∫6−1f(x)dx=−∫−16f(x)dx\int_{6}^{-1} f(x) dx = -\int_{-1}^{6} f(x) dx, we get ∫−3−1f(x)dx=∫−36f(x)dx−∫−16f(x)dx\int_{-3}^{-1} f(x) dx = \int_{-3}^{6} f(x) dx - \int_{-1}^{6} f(x) dx.
That's exactly what we wanted!

STEP 5

We're looking for aa and bb such that ∫absin⁡x dx<0\int_{a}^{b} \sin x \, dx < 0.
Remember, ∫sin⁡x dx=−cos⁡x\int \sin x \, dx = -\cos x.
So, ∫absin⁡x dx=−cos⁡x∣ab=−cos⁡b+cos⁡a\int_{a}^{b} \sin x \, dx = -\cos x \Big|_a^b = -\cos b + \cos a.
We want −cos⁡b+cos⁡a<0-\cos b + \cos a < 0, or cos⁡a<cos⁡b\cos a < \cos b.
One possible solution is a=0a = \mathbf{0} and b=πb = \mathbf{\pi}.

STEP 6

We want ∫abcos⁡x dx=0\int_{a}^{b} \cos x \, dx = 0.
We know ∫cos⁡x dx=sin⁡x\int \cos x \, dx = \sin x.
So, ∫abcos⁡x dx=sin⁡x∣ab=sin⁡b−sin⁡a\int_{a}^{b} \cos x \, dx = \sin x \Big|_a^b = \sin b - \sin a.
We need sin⁡b−sin⁡a=0\sin b - \sin a = 0, or sin⁡a=sin⁡b\sin a = \sin b.
One simple solution is a=0a = \mathbf{0} and b=πb = \mathbf{\pi}.

STEP 7

We need to evaluate ∫01−f(x)dx\int_{0}^{1} -f(x) dx.
This is the negative of the area under the graph of f(x)f(x) from x=0x = 0 to x=1x = 1.
The area is a triangle with base 1 and height 1, so its area is 12⋅1⋅1=12\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}.
Therefore, ∫01−f(x)dx=−12\int_{0}^{1} -f(x) dx = \mathbf{-\frac{1}{2}}.

STEP 8

We need ∫110f(x)dx\int_{11}^{0} f(x) dx.
This is the negative of ∫011f(x)dx\int_{0}^{11} f(x) dx.
We can break this integral into geometric shapes.
From x=0x = 0 to x=3x = 3, we have a triangle with area 12⋅3⋅2=3\frac{1}{2} \cdot 3 \cdot 2 = 3.
From x=3x = 3 to x=4x = 4, we have a rectangle with area 1⋅2=21 \cdot 2 = 2.
From x=4x = 4 to x=8x = 8, we have a triangle with area 12⋅4⋅4=8\frac{1}{2} \cdot 4 \cdot 4 = 8.
From x=8x = 8 to x=11x = 11, we have a triangle with area 12⋅3⋅3=92\frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}.
The total area is 3+2−8+92=123 + 2 - 8 + \frac{9}{2} = \frac{1}{2}.
So, ∫110f(x)dx=−12\int_{11}^{0} f(x) dx = \mathbf{-\frac{1}{2}}.

STEP 9

We have ∫34(6−f(x))dx=∫346dx−∫34f(x)dx\int_{3}^{4} (6 - f(x)) dx = \int_{3}^{4} 6 dx - \int_{3}^{4} f(x) dx.
The first integral is 6x∣34=6(4−3)=66x \Big|_3^4 = 6(4 - 3) = 6.
The second integral is the area of a rectangle with width 1 and height 2, which is 2.
So, ∫34(6−f(x))dx=6−2=4\int_{3}^{4} (6 - f(x)) dx = 6 - 2 = \mathbf{4}.

STEP 10

We have ∫1043f(x)dx=−3∫410f(x)dx\int_{10}^{4} 3f(x) dx = -3 \int_{4}^{10} f(x) dx.
From x=4x = 4 to x=8x = 8, the area is −8-8, and from x=8x = 8 to x=10x = 10, the area is 12⋅2⋅2=2\frac{1}{2} \cdot 2 \cdot 2 = 2.
So, ∫410f(x)dx=−8+2=−6\int_{4}^{10} f(x) dx = -8 + 2 = -6.
Therefore, ∫1043f(x)dx=−3(−6)=18\int_{10}^{4} 3f(x) dx = -3(-6) = \mathbf{18}.

STEP 11

a. a=−2a = -2, b=5b = 5 b. a=−3a = -3, b=−1b = -1 c. a=0a = 0, b=πb = \pi d. a=0a = 0, b=πb = \pi A. −12-\frac{1}{2} B. −12-\frac{1}{2} C. 4 D. 18

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