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mathematics
precalculus
Thomas Calculus Early Transcendentals 13th Edition Joel R Hass, Christopher E Heil, Maurice D Weir - Solutions
Use the Substitution Formula in Theorem 7 to evaluate the integral. S (y³ + 6y² − 12y + 9)−1/2 (y² + 4y − 4) dy 0
Graph the integrands and use known area formulas to evaluate the integrals. -1 xp (x-7) of
Graph the integrands and use known area formulas to evaluate the integrals. La -1 (1 − x) dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. -1/2 -1 t² sin² 1 + ¦) ₁ t dt
Use the Substitution Formula in Theorem 7 to evaluate the integral. √√7² 2 Jo Ve cos² (0³/2) de
Graph the integrands and use known area formulas to evaluate the integrals. Lavida (1 + V1 - x²) dx -1
Use known area formulas to evaluate the integral. .b 4x dx, b>0 0
Use known area formulas to evaluate the integral. X [²2/dx, b>0 lx. 0
Use the Substitution Formula in Theorem 7 to evaluate the integral. 0 π/4 (1 + etan) sec²0 de
Use the Substitution Formula in Theorem 7 to evaluate the integral. TT So 0 2 sin t COS t dt
Use the Substitution Formula in Theorem 7 to evaluate the integral. TT/2 π/4 (1 + ecot) csc²0 de
Use known area formulas to evaluate the integral. a 3t dt, 0< a
Use the results of Equations (2) and (4) to evaluate the integral. 1 √2 x dx
Use known area formulas to evaluate the integral. b [²2x6 a 2s ds, 0 < a
Use the results of Equations (2) and (4) to evaluate the integral. 27 TT Ꮎ dᎾ
Use known area formulas to evaluate the integral.on a. [-2, 2], b. [0, 2] f(x) = √4x²
Use the results of Equations (2) and (4) to evaluate the integral. 5√2 V2 r dr
Use the Substitution Formula in Theorem 7 to evaluate the integral. 0 T/3 4 sin 0 1 - 4 cos 0 de
Use known area formulas to evaluate the integral.on a. [-1, 0], b. [-1, 1] IA f(x) = 3x + V1 - x²
Use the Substitution Formula in Theorem 7 to evaluate the integral. S 1 2 ln x X dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. 2 4 dx x ln x
Use the results of Equations (2) and (4) to evaluate the integral. 0 0.3 S s² ds
Use the results of Equations (2) and (4) to evaluate the integral. JO 3/7 x² dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. 4 S 2 dx x (In x)²
Use the Substitution Formula in Theorem 7 to evaluate the integral. 16 dx √In x 2 2x1 2x
Use the Substitution Formula in Theorem 7 to evaluate the integral. 0 TT/2 X 12 dx tan
Use the results of Equations (2) and (4) to evaluate the integral. 0 1/2 t² dt
Use the Substitution Formula in Theorem 7 to evaluate the integral. TT/3 S 0 tan² 0 cos 0 de
Use the Substitution Formula in Theorem 7 to evaluate the integral. TT/2 π/4 cot t dt
Use the results of Equations (2) and (4) to evaluate the integral. 0 11 / 2 Ꮎ- dᎾ
Use the Substitution Formula in Theorem 7 to evaluate the integral. 0 7/12 6 tan 3x dx
Use the results of Equations (2) and (4) to evaluate the integral. a 2a x dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. 7/2 2 cos 0 de 7/21 + (sin 0)²
Use the results of Equations (2) and (4) to evaluate the integral. a За x dx
Guess an antiderivative for the integrand function. Validate your guess by differentiation and then evaluate the given definite integral. S 1 In x X dx
Guess an antiderivative for the integrand function. Validate your guess by differentiation and then evaluate the given definite integral. L'x 0 Xer dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. T/4 TT/6 csc² x dx 1 + (cot.x)²
Use the results of Equations (2) and (4) to evaluate the integral. Jo b x² dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. 0 In √3 ex dx 1 + ²x X
Guess an antiderivative for the integrand function. Validate your guess by differentiation and then evaluate the given definite integral. TT/3 sin² x cos x dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Guess an antiderivative for the integrand function. Validate your guess by differentiation and then evaluate the given definite integral. 2 5 x dx V1 + x²
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. 0 √2 (t-√2) d dt
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. [121 0 (2t - 3) dt
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. S 3 7 dx
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. 0 2 5x dx
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. T 3 (2z - 3) dz
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. 2 1 + dz
Use the Substitution Formula in Theorem 7 to evaluate the integral. THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, b] and f is continuous on the range of g(x) = u, then rg(b) [ f(g(x)) g'(x) dx = - f(u) du. g(a) Proof Let F denote any antiderivative of f.
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. 1/2 24u² du
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. Г 2 3u² du
Find the total areas of the shaded region. y=x√4x² -2 0 2 X
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. .0 [ (31² 1 (3x² + x - 5) dx
Use the rules in Table 5.6 and Equations (2)–(4) to evaluate the integral. 0 2 (3x² + x 5) dx -
Find the total areas of the shaded region. y = (cos x)(sin(7 + 7 sin x)) y π -1 2 0 -¹
Find the total areas of the shaded region. y y = (1 - cos x) sin x TT →x
Find the total areas of the shaded region. y = πT -2 -1 y 3(sin x)/1+ cos x -2 -3 0
Find the total areas of the shaded region. Elm 3 y 2 1 /0 -41 y = 1/scc²t ادرا t 3 y = -4 sin²t
Find the total areas of the shaded region. (-2,8) -2 y 8 y = (2,8) 2x² NOT TO SCALE y=x²- 2x² 2 X
Find the total areas of the shaded region. y y = 1 TT 2 = COS X cos2 y = 77 X
Use a definite integral to find the area of the region between the given curve and the x-axis on the interval [0, b]. y || X +1
Graph the function and find its average value over the given interval. f(x) = 2 on [0,3]
Find the total areas of the shaded region. -1 y 0 y = y= = -2x4 -2-- 1 X
Find the total areas of the shaded region. y 0 x = y3 x = y² (1, 1) 1 X
Find the total areas of the shaded region. x = 2y²-2y 0 X = 1 12y² - 12y3 X
Find the total areas of the shaded region. 1 y 0 y = x 1 y = 1 y = 2 4 X
Use a definite integral to find the area of the region between the given curve and the x-axis on the interval [0, b].y = πx2
Find the total areas of the shaded region. (-3,5) y 5 가 | y = x2 - 4 -3 (-3, -3) ON 1 y = -41 X −x2 - 2x -x² - (1, -3)
Find the total areas of the shaded region. 0 y = x² 1 x+y=2 2 X
Find the total areas of the shaded region. -2 +1 y y = −x2 + 3.x (2, 2) # 1 2 y = 2x3 - x2 - 5x (-2,-10) --- X
Graph the function and find its average value over the given interval.ƒ(x) = x2 - 1 on [0, 23]
Use the method of Example 4a or Equation (1) to evaluate the definite integral.Example 4aCompute ∫0bx dx and find the area A under y = x over the interval [0, b], b > 0. To compute the definite integral as the limit of Riemann sums, we calculate lim-0 =1f(c) Ax for partitions whose norms go to
Each of the following functions solves one of the initial value problem. Which function solves which problem? Give brief reasons for your answers.a.b.c.d. y = S 1 1 -dt lt - 3
Graph the function and find its average value over the given interval.ƒ(x) = -3x2 - 1 on [0, 1]
Use the method of Example 4a or Equation (1) to evaluate the definite integral.Example 4aCompute ∫0bx dx and find the area A under y = x over the interval [0, b], b > 0. To compute the definite integral as the limit of Riemann sums, we calculate lim-0 =1f(c) Ax for partitions whose norms go to
Each of the following functions solves one of the initial value problem. Which function solves which problem? Give brief reasons for your answers.a.b.c.d. y = S 1 1 -dt lt - 3
Find the total area between the region and the x-axis.y = -x2 - 2x, -3 ≤ x ≤ 2
Graph the function and find its average value over the given interval.ƒ(x) = 3x2 - 3 on [0, 1]
Find the total areas of the shaded region. (-2,4) -2-1 y 2 -5-- y=4-x² 1 2 3 X y = -x + 2 (3,-5)
Find the total areas of the shaded region. y 3 -2, (3,6) X 3 y = (3, 1) 3 8/3 X
Use the method of Example 4a or Equation (1) to evaluate the definite integral.Example 4aCompute ∫0bx dx and find the area A under y = x over the interval [0, b], b > 0. To compute the definite integral as the limit of Riemann sums, we calculate lim-0 =1f(c) Ax for partitions whose norms go to
Find the areas of the shaded region. y 2 0 y = 2 y = 1 + cos x X = TT ㅠ
Find the total area between the region and the x-axis.y = 3x2 - 3, -2 ≤ x ≤ 2
Graph the function and find its average value over the given interval.ƒ(t) = (t - 1)2 on [0, 3]
Use the method of Example 4a or Equation (1) to evaluate the definite integral.Example 4aCompute ∫0bx dx and find the area A under y = x over the interval [0, b], b > 0. To compute the definite integral as the limit of Riemann sums, we calculate lim-0 =1f(c) Ax for partitions whose norms go to
Find the areas of the shaded region. TT 4 y √2 -√₂ 0 y = sec 0 tan 0 TT 4 0
Find the total area between the region and the x-axis.y = x3 - 3x2 + 2x, 0 ≤ x ≤ 2
Find the areas of the shaded region. У TT 6 y = sin x 5 п 6 -Х
Graph the function and find its average value over the given interval.ƒ(t) = t2 - t on [-2, 1]
Use the method of Example 4a or Equation (1) to evaluate the definite integral.Example 4aCompute ∫0bx dx and find the area A under y = x over the interval [0, b], b > 0. To compute the definite integral as the limit of Riemann sums, we calculate lim-0 =1f(c) Ax for partitions whose norms go to
Find the total area between the region and the x-axis.y = x1/3 - x, -1 ≤ x ≤ 8
Graph the function and find its average value over the given interval.g(x) = |x| - 1 on a. [-1, 1], b. [1, 3], and c. [-1, 3]
Find the areas of the shaded region. y = sec² t 4 y 2 y = 1 - 1² 0 1
Graph the function and find its average value over the given interval.h(x) = -|x| on a. [-1, 0]b. [0, 1]c. [-1, 1]
If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitution.a. u = tan x, followed by y = v3, then by w =
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