Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍
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Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍
Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).
(a) Find and graph the area function A (𝓍) = ∫ₐˣ ƒ(t) dt .
ƒ(t) = 2t + 5 , a = 0
Working with area functions Consider the function ƒ and its graph.
(a) Estimate the zeros of the area function A(𝓍) = ∫₀ˣ ƒ(t) dt , for 0 ≤ 𝓍 ≤ 10 .
Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₋₂ˣ ƒ(t) dt and F(x) = ∫₄ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.
(a) A (―2)
Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1
(a) ∫₀^π/2 (2 sin θ ― cos θ) dθ
Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.
(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)