Video Transcript
The power of an engine at time 𝑡
seconds is given by 𝑃 of 𝑡 is equal to 12𝑡 squared plus eight 𝑡 watts. Find the work done by the engine
between 𝑡 equals two seconds and 𝑡 equals three seconds.
In this question, we are given an
expression for the power of an engine in terms of 𝑡. We recall that power is the
derivative of work such that 𝑃 is equal to d𝑊 by d𝑡. As integration is the opposite or
inverse of differentiation, the work must be the integral of the power with respect
to 𝑡. We can therefore find an expression
for the work done by integrating 12𝑡 squared plus eight 𝑡 with respect to 𝑡. As we need to find the work done
between 𝑡 equals two and 𝑡 equals three seconds, our lower and upper limits will
be two and three, respectively. This can also be demonstrated
graphically as shown. The work done will be equal to the
area bounded by the curve the vertical lines 𝑡 equals two and 𝑡 equals three and
the horizontal axis.
We can integrate the expression for
power term by term. We recall that the integral of 𝑎𝑥
to the power of 𝑛 with respect to 𝑥 is equal to 𝑎𝑥 to the power of 𝑛 plus one
divided by 𝑛 plus one where 𝑛 is not equal to negative one. In this question, since we’re
dealing with a definite integral, we will not need to include the constant 𝐶. Integrating 12𝑡 squared gives us
12𝑡 cubed over three, which simplifies to four 𝑡 cubed. Integrating eight 𝑡 gives us eight
𝑡 squared over two, which simplifies to four 𝑡 squared.
Our next step is to substitute our
upper and lower limits. This gives us four multiplied by
three cubed plus four multiplied by three squared minus four multiplied by two cubed
plus four multiplied by two squared. This in turn simplifies to 144
minus 48, which is equal to 96. Since the power of the engine was
measured in the standard unit of watts, the work will be measured in joules. The work done by the engine between
𝑡 equals two and 𝑡 equals three seconds is 96 joules.
