In this lesson, we will use
the free online graphing calculator, linked below,
to graph integrand functions,
and evaluate definite integrals.
We'll begin by evaluating
the definite integrals on the left,
with the integrand function is,
f of x equals negative 1/10 X squared, plus eight.
Let's go to the graphing calculator,
and enter the integrand function.
Let's first clear the existing functions,
under f of X, we enter negative 1/10 X squared plus eight.
Remember the character exponent key
is Shift+ Command + 6, from the keyboard.
And now we click plot graph.
Notice how we don't see the graph
on the coordinate plane,
because we need to adjust the window.
And we do need to adjust the window,
because in order to evaluate a definite integral,
we use the calculation menu under the graphing calculator,
not the calculation window below.
To adjust the window,
let's click the minus,
at the bottom right hand corner, to zoom out.
Notice now we do see the graph of the function.
So we could stop here,
but since the limits of integration
are always greater than or equal to zero,
let's adjust the window manually,
by clicking on the window tab.
And let's change the x-min to negative two,
the x-max to 10, the y-min to negative two,
and the y-max to 10.
Now we have a better view of our graph,
for our definite integrals.
So let's evaluate the first definite integral,
with limits of integration are from zero to seven.
We click the calculation tab, click the drop down menu,
and select integral.
Next, click on the curve.
Enter the limits of integration,
which are from zero to seven.
Click calculate.
The definite integrals approximately 44.5667,
because the function is non-negative
of interval from zero to seven,
this value does represent the area bounded
by the function on the x-axis,
over this interval.
The second integral, has a lower limit
of integration of square root three.
We go back change the zero to square root three SQRT
with a three in parentheses,
the upper limit of integration,
is two square root seven.
Click calculate.
The value of the definite integral,
is approximate 23.7101.
And again, the function is non-negative over this interval,
and therefore this does represent
the area, bounded by the function over this interval.
Let's go ahead and record these values.
And now to evaluate the second definite integrals,
where the integrand function is three sine x.
For the first integral,
the limits of integration are from zero to pi.
And for the second, the limits are from zero to two pi.
So going back to the calculator,
one thing I did notice,
is to evaluate another integral,
we either need to refresh the calculator and start over,
or click a different function under mode,
and then come back and select integral.
I'm going to go ahead and just refresh the calculator,
by clicking the refresh button on the browser.
Clear out the functions into the new function,
of three sine x under f of x.
Click plot graph.
The graph does not show the X interval from zero two pi.
So let's go ahead...
which is approximately from zero to 6.28.
So let's adjust the window manually,
by clicking window.
And let's change the x-min to let's say negative one,
the x-max to seven and click update.
And now to evaluate the definite integral
from zero to pi, click calculate.
Click the drop down menu, select integral,
click on the curve, into the limits
of integration from zero to pi,
enter p-i four pi, and click calculate.
The interval from zero to pi,
is this interval here from zero,
to approximately 3.14,
where the function is not negative.
And therefore the area below the function,
and above the x-axis over this interval,
is exactly equal to six square units.
The last integral, the limits of integration
are from zero to two pi,
so we change pi to two pi,
and click calculate.
Notice this integral is equal to zero.
And that's because we can think of,
the value of the definite integral as signed area.
One of the areas above the x-axis, is positive,
and the areas below the x-axis, it's negative.
So the interval from zero to two pi,
is approximately from zero to 6.28.
From this point on the left to this point on the right.
The area above the x-axis is positive,
the area below the x-axis is negative,
and because the two areas are the same but opposite sign,
the sum is equal to zero,
which is why the value of the definite integral is zero.
Let's go ahead and record these values
of the definite integrals.
I hope you found this helpful.