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This section explores the concept of antiderivatives and their relation to slope fields. Utilizing the power rule, we find the indefinite integral of a polynomial function ∫(4t³ + 6t² - 5) dt, leading to t⁴ + 2t³ - 5t + C. We also discuss how antiderivatives signify functions with specific slopes, noting that multiple functions share the same slope form but differ by a constant. An example is provided involving initial conditions to determine the integration constant C, with further applications of integrals highlighting properties such as linearity.
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The Indefinite Integral Find: ∫4t3+6t2-5 dt ∫axn dx=(axn+1)/(n+1) Use the power rule in reverse. t4+2t3-5t Here is your antiderivative. There is one more step. Like a derivative represents a slope of a given function, an antiderivative represents the function with a given slope. Because the slope of a function tells us nothing about its location, an antiderivative can actually represent infinitely many functions that are essentially the same curve shifted up or down. This is represented by adding an integration constant “+C” to the end of an indefinite integral. ∫4t3+6t2-5dt = t4+2t3-5t+C
Integrating With an Initial Condition Find: ∫4t3+6t2-5dt Given: f(1) = 0 ∫4t3+6t2-5dt = t4+2t3-5t+C Begin with the indefinite integral (taken from previous example) Because we are given a value of the function at a certain value of t, we can determine the value of C by plugging the numbers into the equation. 0 = 14+2(1)3-5(1)+C Substitute 1 for t and set equal to 0. C=2 Solve for C. • f(t) = t4+2t3-5t+2 Plug C into the original equation.
Applying some Properties of Integrals Find: ∫2f(x) - 3g(x) dx Given: ∫f(x) = 4x and ∫g(x) = 3x • ∫2f(x) dx -∫3g(x)dx Split into two integrals. • 2∫f(x) dx -3∫g(x)dx Pull out the constants. • 2(4x)- 3(3x) Plug in the given values. ∫2f(x) - 3g(x) dx = -x Simplify.