Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problema 6.RE.3d

Capitolo 6, Problema 6.RE.3d

2–3. Displacement, distance, and position Consider an object moving along a line with the following velocities and initial positions. Assume time t is measured in seconds and velocities have units of m/s.

d. Determine the position function s(t) using the Fundamental Theorem of Calculus (Theorem 6.1). Check your answer by finding the position function using the antiderivative method.

v(t) = 12t²-30t+12, for 0 ≤ t ≤ 3; s(0)=1

Guida verificata passo dopo passo

Recall that the position function $s(t)$ can be found from the velocity function $v(t)$ by integrating $v(t)$ over time and adding the initial position $s(0)$. According to the Fundamental Theorem of Calculus, we have: \[s(t) = s(0) + \int_0^t v(x) \, dx\]

Substitute the given velocity function $v(t) = 12t^2 - 30t + 12$ into the integral: \[s(t) = 1 + \int_0^t (12x^2 - 30x + 12) \, dx\]

Compute the integral by integrating each term separately: - The integral of \$12x^2$ is $12 \cdot \frac{x^3}{3} = 4x^3$ - The integral of $-30x$ is $-30 \cdot \frac{x^2}{2} = -15x^2$ - The integral of $12$ is \$12x$

Combine the results of the integrals and evaluate the definite integral from 0 to $t$: \[\int_0^t (12x^2 - 30x + 12) \, dx = \left[4x^3 - 15x^2 + 12x\right]_0^t = 4t^3 - 15t^2 + 12t\]

Add the initial position $s(0) = 1$ to the integral result to write the position function: \[s(t) = 1 + 4t^3 - 15t^2 + 12t\] To check your answer using the antiderivative method, differentiate $s(t)$ and verify that $s'(t) = v(t)$.

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Fundamental Theorem of Calculus

This theorem links differentiation and integration, stating that the integral of a function's derivative over an interval gives the net change in the function's values. In this problem, it allows us to find the position function s(t) by integrating the velocity function v(t) over time.

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Fundamental Theorem of Calculus Part 1

Antiderivative Method

Finding an antiderivative involves determining a function whose derivative is the given function. Here, integrating v(t) = 12t² - 30t + 12 gives s(t) plus a constant, which can be found using the initial position s(0) = 1.

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Antiderivatives

Initial Conditions in Differential Equations

Initial conditions specify the value of a function at a particular point, enabling the determination of the constant of integration after finding an antiderivative. Using s(0) = 1 ensures the position function s(t) accurately reflects the object's starting position.

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Solutions to Basic Differential Equations

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