Chapter 7.1, Problem 69E

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Chapter
Section

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

# A particle that moves along a straight line has velocity v ( t ) = t 2 e − t meters per second after t seconds. How far will it travel during the first t seconds?

To determine

To find: distance covered by a particle moving in straight line during its first t seconds.

Explanation

Recall that velocity is the rate at which the position of an object changes. It is given by the ratio of distance traveled to the time taken in covering that distance. Then, the distance covered will be the product of time and velocity.

Formula used:

If v(t) is the velocity which changes with time, then distance traveled in time t will be given by the following integral

d=0tv(t)dt

Given:

Velocity of the particle at time t, v(t)=t2et

Calculation:

Substitute the expression for velocity v in the formula for distance

d=0tt2etdt

Solve the integral using integration by parts. The technique of integration by parts comes in handy when the integrand involves product of two functions. It can be thought of as a rule corresponding to the product rule in differentiation. The formula for integration by parts for definite integral is given by

abf(x)g(x)dx=f(x)g(x)]ababg(x)f(x)dx

Make the choice for u and dv such that the resulting integration from the formula above is easier to integrate. Let

u=t2      dv=etdt

Then, the differentiation of u and antiderivative of dv will be

du=2tdt      v=et

Substitute for variables in the formula above to get

0tt2etdt=t2(et)]0t0t(et)2tdt=t2(et)]0t+20ttetdt …… (1)

Solve the integral in the last term by again applying integration by parts with variables as:

u=t      dv=etdt

Then, the differentiation of u and antiderivative of dv will be

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