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∫−4016−x2dx\int_{-4}^{0} \sqrt{16-x^{2}} d x
Mar 10, 2024
∫−4016−x2dx\int_{-4}^{0} \sqrt{16-x^{2}} d x
Generated Graph
Solution by Steps
step 1
To solve the integral ∫−4016−x2dx\int_{-4}^{0} \sqrt{16-x^{2}} dx, we recognize that it represents the area of a quarter circle with radius 4
step 2
The area of a full circle with radius 4 is πr2=π⋅42=16π\pi r^{2} = \pi \cdot 4^{2} = 16\pi
step 3
Since we want the area of a quarter circle, we divide the full circle's area by 4, which gives us 16π4=4π\frac{16\pi}{4} = 4\pi
step 4
The definite integral from -4 to 0 of the function 16−x2\sqrt{16-x^{2}} is equal to the area of the quarter circle, which is 4π4\pi
Answer
∫−4016−x2dx=4π\int_{-4}^{0} \sqrt{16-x^{2}} dx = 4\pi
Key Concept
Area under a curve represented by an integral
Explanation
The integral of the square root of a squared radius minus the variable squared from -radius to 0 represents the area of a quarter circle.
Suppose that ff and hh are continuous functions and that ∫19f(x)dx=−1,∫79f(x)dx=5,∫79h(x)dx=4. \int_{1}^{9} f(x) d x=-1, \quad \int_{7}^{9} f(x) d x=5, \quad \int_{7}^{9} h(x) d x=4 .
Solution by Steps
step 1
Use the properties of definite integrals to find the value of ∫17f(x)dx\int_{1}^{7} f(x) dx
step 2
We know that ∫19f(x)dx=∫17f(x)dx+∫79f(x)dx\int_{1}^{9} f(x) dx = \int_{1}^{7} f(x) dx + \int_{7}^{9} f(x) dx
step 3
Substitute the given values to find ∫17f(x)dx\int_{1}^{7} f(x) dx
step 4
We have −1=∫17f(x)dx+5-1 = \int_{1}^{7} f(x) dx + 5
step 5
Solving for ∫17f(x)dx\int_{1}^{7} f(x) dx gives us ∫17f(x)dx=−1−5\int_{1}^{7} f(x) dx = -1 - 5
step 6
Therefore, ∫17f(x)dx=−6\int_{1}^{7} f(x) dx = -6
Answer
∫17f(x)dx=−6\int_{1}^{7} f(x) dx = -6
Key Concept
Properties of Definite Integrals
Explanation
The integral of a function over an interval can be split into the sum of integrals over adjacent subintervals.
Solution by Steps
step 1
Use the given value of ∫79h(x)dx\int_{7}^{9} h(x) dx to find ∫17h(x)dx\int_{1}^{7} h(x) dx
step 2
We need to find the value of ∫19h(x)dx\int_{1}^{9} h(x) dx, but we only have the value from 7 to 9
step 3
Since we don't have enough information to find ∫17h(x)dx\int_{1}^{7} h(x) dx, we cannot determine ∫19h(x)dx\int_{1}^{9} h(x) dx
Answer
Insufficient information to determine ∫17h(x)dx\int_{1}^{7} h(x) dx
Key Concept
Definite Integral over an Interval
Explanation
Without additional information about the function h(x) over the interval from 1 to 7, we cannot calculate the integral.
The exercises in this section are designed to reinforce your understanding of the definite integral from the algebraic and geometric points of view. For this reason, you should not use the numerical integration capability of your calculator (NINT) except perhaps to support an answer. 1. Suppose that ff and gg are continuous functions and that ∫12f(x)dx=−4,∫15f(x)dx=6,∫15g(x)dx=8. \int_{1}^{2} f(x) d x=-4, \int_{1}^{5} f(x) d x=6, \int_{1}^{5} g(x) d x=8 . Use the rules in Table 6.3 to find each integral. (a) ∫22g(x)dx\int_{2}^{2} g(x) d x (b) ∫51g(x)dx\int_{5}^{1} g(x) d x (c) ∫123f(x)dx\int_{1}^{2} 3 f(x) d x (d) ∫25f(x)dx\int_{2}^{5} f(x) d x (e) ∫15[f(x)−g(x)]dx\int_{1}^{5}[f(x)-g(x)] d x (f) ∫15[4f(x)−g(x)]dx\int_{1}^{5}[4 f(x)-g(x)] d x 2. Suppose that ff and hh are continuous functions and that ∫19f(x)dx=−1,∫79f(x)dx=5,∫79h(x)dx=4 \int_{1}^{9} f(x) d x=-1, \quad \int_{7}^{9} f(x) d x=5, \quad \int_{7}^{9} h(x) d x=4 Use the rules in Table 6.3 to find each integral. (a) ∫19−2f(x)dx\int_{1}^{9}-2 f(x) d x (b) ∫79[f(x)+h(x)]dx\int_{7}^{9}[f(x)+h(x)] d x (c) ∫79[2f(x)−3h(x)]dx\int_{7}^{9}[2 f(x)-3 h(x)] d x (d) ∫91f(x)dx\int_{9}^{1} f(x) d x (e) ∫17f(x)dx\int_{1}^{7} f(x) d x (f) ∫97[h(x)−f(x)]dx\int_{9}^{7}[h(x)-f(x)] d x 3. Suppose that ∫12f(x)dx=5\int_{1}^{2} f(x) d x=5. Find each integral. (a) ∫12f(u)du\int_{1}^{2} f(u) d u (b) ∫123f(z)dz\int_{1}^{2} \sqrt{3} f(z) d z (c) ∫21f(t)dt\int_{2}^{1} f(t) d t (d) ∫12[−f(x)]dx\int_{1}^{2}[-f(x)] d x 4. Suppose that ∫−30g(t)dt=2\int_{-3}^{0} g(t) d t=\sqrt{2}. Find each integral. (a) ∫0−3g(t)dt\int_{0}^{-3} g(t) d t (b) ∫−30g(u)du\int_{-3}^{0} g(u) d u (c) ∫−30[−g(x)]dx\int_{-3}^{0}[-g(x)] d x (d) ∫−30g(r)2dr\int_{-3}^{0} \frac{g(r)}{\sqrt{2}} d r Section 6.3 Definite Integrals and Antiderivatives 299 5. Suppose that ff is continuous and that ∫03f(z)dz=3 and ∫04f(z)dz=7. \int_{0}^{3} f(z) d z=3 \text { and } \int_{0}^{4} f(z) d z=7 . Find each integral. (a) ∫34f(z)dz\int_{3}^{4} f(z) d z (b) ∫43f(t)dt\int_{4}^{3} f(t) d t 6. Suppose that hh is continuous and that ∫−11h(r)dr=0 and ∫−13h(r)dr=6. \int_{-1}^{1} h(r) d r=0 \text { and } \int_{-1}^{3} h(r) d r=6 . Find each integral. (a) ∫13h(r)dr\int_{1}^{3} h(r) d r (b) −∫31h(u)du-\int_{3}^{1} h(u) d u 7. Show that the value of ∫01sin⁡(x2)dx\int_{0}^{1} \sin \left(x^{2}\right) d x cannot possibly be 2 . 8. Show that the value of ∫01x+8dx\int_{0}^{1} \sqrt{x+8} d x lies between 22≈2.82 \sqrt{2} \approx 2.8 and 3 . 9. Integrals of Nonnegative Functions Use the Max-Min Inequality to show that if ff is integrable then f(x)≥0 on [a,b]⇒∫abf(x)dx≥0. f(x) \geq 0 \text { on }[a, b] \Rightarrow \int_{a}^{b} f(x) d x \geq 0 . 10. Integrals of Nonpositive Functions Show that if ff is integrable then f(x)≤0 on [a,b]⇒∫abf(x)dx≤0. f(x) \leq 0 \text { on }[a, b] \Rightarrow \int_{a}^{b} f(x) d x \leq 0 . In Exercises 11-14, use NINT to find the average value of the function on the interval. At what point(s) in the interval does the function assume its average value? 11. y=x2−1y=x^{2}-1, [0,3][0, \sqrt{3}] 12. y=−x22y=-\frac{x^{2}}{2}. [0,3][0,3] 13. y=−3x2−1,[0,1]y=-3 x^{2}-1,[0,1] 14. y=(x−1)2y=(x-1)^{2}, 17. f(t)=sin⁡t,[0,2π]f(t)=\sin t,[0,2 \pi] 18. f(θ)=tan⁡θ,[−π4,π4]f(\theta)=\tan \theta,\left[-\frac{\pi}{4}, \frac{\pi}{4}\right] In Exercises 19-30, interpret the integrand as the rate of change of a quantity and evaluate the integral using the antiderivative of the quantity, as in Example 4. 19. ∫π2πsin⁡xdx\int_{\pi}^{2 \pi} \sin x d x 20. ∫0π/2cos⁡xdx\int_{0}^{\pi / 2} \cos x d x 21. ∫01exdx\int_{0}^{1} e^{x} d x 22. ∫0π/4sec⁡2xdx\int_{0}^{\pi / 4} \sec ^{2} x d x 23. ∫142xdx\int_{1}^{4} 2 x d x 24. ∫−123x2dx\int_{-1}^{2} 3 x^{2} d x 25. ∫−265dx\int_{-2}^{6} 5 d x 26. ∫378dx\int_{3}^{7} 8 d x 27. ∫−1111+x2dx\int_{-1}^{1} \frac{1}{1+x^{2}} d x 28. ∫01/211−x2dx\int_{0}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} d x 29. ∫1e1xdx\int_{1}^{e} \frac{1}{x} d x 30. ∫14−x−2dx\int_{1}^{4}-x^{-2} d x In Exercises 31-36, find the average value of the function on the interval, using antiderivatives to compute the integral. 31. y=sin⁡x,[0,π]y=\sin x, \quad[0, \pi] 32. y=1x,[e,2e]y=\frac{1}{x}, \quad[e, 2 e] 33. y=sec⁡2x,[0,π4]y=\sec ^{2} x, \quad\left[0, \frac{\pi}{4}\right] 34. y=11+x2,[0,1]y=\frac{1}{1+x^{2}}, \quad[0,1] 35. y=3x2+2x,[−1,2]y=3 x^{2}+2 x, \quad[-1,2] 36. y=sec⁡xtan⁡x,[0,π3]y=\sec x \tan x, \quad\left[0, \frac{\pi}{3}\right] 37. Group Activity Use the Max-Min Inequality to find upper and lower bounds for the value of
Generated Graph
Solution by Steps
step 1
To find the integral of a constant function over an interval where the upper and lower limits are the same, the result is always 0
step 2
Applying this to ∫22g(x)dx \int_{2}^{2} g(x) dx , we get 0 since the limits are the same
step 3
To find the integral of a function over an interval [a, b], and then from [b, a], the results are negatives of each other
step 4
Since ∫15g(x)dx=8 \int_{1}^{5} g(x) dx = 8 , reversing the limits, ∫51g(x)dx=−8 \int_{5}^{1} g(x) dx = -8
step 5
To find the integral of a constant multiplied by a function, we multiply the constant by the integral of the function over the interval
step 6
Given ∫12f(x)dx=−4 \int_{1}^{2} f(x) dx = -4 , then ∫123f(x)dx=3×−4=−12 \int_{1}^{2} 3f(x) dx = 3 \times -4 = -12
step 7
The integral of a function over an interval [a, c] can be found by adding the integrals from [a, b] and [b, c]
step 8
Given ∫12f(x)dx=−4 \int_{1}^{2} f(x) dx = -4 and ∫15f(x)dx=6 \int_{1}^{5} f(x) dx = 6 , then ∫25f(x)dx=6−(−4)=10 \int_{2}^{5} f(x) dx = 6 - (-4) = 10
step 9
To find the integral of the difference of two functions, subtract the integral of the second function from the integral of the first function over the same interval
step 10
Given ∫15f(x)dx=6 \int_{1}^{5} f(x) dx = 6 and ∫15g(x)dx=8 \int_{1}^{5} g(x) dx = 8 , then ∫15[f(x)−g(x)]dx=6−8=−2 \int_{1}^{5} [f(x) - g(x)] dx = 6 - 8 = -2
step 11
To find the integral of a linear combination of functions, distribute the integral over the sum and multiply by the constants
step 12
Given ∫15f(x)dx=6 \int_{1}^{5} f(x) dx = 6 and ∫15g(x)dx=8 \int_{1}^{5} g(x) dx = 8 , then ∫15[4f(x)−g(x)]dx=4×6−8=24−8=16 \int_{1}^{5} [4f(x) - g(x)] dx = 4 \times 6 - 8 = 24 - 8 = 16
Answer
(a) 0 (b) -8 (c) -12 (d) 10 (e) -2 (f) 16
Key Concept
Properties of definite integrals, including the use of additive intervals and the effect of constant multipliers.
Explanation
The solutions utilize the given integral values and apply the properties of definite integrals, such as the zero integral over a zero-width interval, the negation when reversing limits, and the linearity of integration.
Please note that the above response only addresses the first set of integrals (1a to 1f) from the student's question. If you need the solutions for the remaining parts, please let me know, and I will provide them following the same format.
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