All Calculus 1 Resources
Example Questions
Example Question #61 : How To Find Area Of A Region
Integrate
Let
Therefore by u substitution we are able to integrate thus finding the area under the curve. Once you have integrated remember to plug your original variables back into the equation:
Example Question #61 : How To Find Area Of A Region
Find the area between the curves and .
Looking at the graph, we see that the graphs intersect at the point with on top.
Therefore, our integral to find the area by partitioning the x-axis looks like
.
Example Question #61 : Area
Find the area between and the -axis.
First we find that the equation crosses the y-axis at y=-1, 1.
Partitioning the y-axis, we get the area is equal to
.
Example Question #4056 : Calculus
Calculate the area between the sine curve and the cosine curve for
.
To find the area between the 2 trigonometric waves, we have to look at two different regions between 0 and .
Cosine is above sine from 0 to , and sine is above cosine from to :
Written as an integral, this area can be represented as follows:
Integrating yields:
from 0 to and from to
Evaluate:
rationalizing the denominator:
Example Question #4057 : Calculus
Find the area between the functions and .
The area between the two curves goes between the two points of intersection:
To figure out the left and right bounds of this region, solve the system of equations by setting the two equations equal to each other and solving for x:
add to both sides
subtract 1 from both sides
divide by 4
The solutions are -1 and 1, so we're finding the area for .
For this region, the function is above , so our integral looks like this:
We can simplify before solving to and continue solving.
from -1 to 1
Example Question #3022 : Functions
Find the area between and .
First, figure out the boundaries of the region between the two curves. To do this, set the two functions equal to each other to determine where the curves intersect:
multiply both sides by x
This can be re-written in more standard quadratic form:
This can be factored as , so our two solutions are 2 and 3. We are then evaluating the definite integral between 2 and 3. In this region, the equation is greater. Evaluate:
between 2 and 3
Example Question #4059 : Calculus
Find the dot product and .
None of these
The way to find a dot product is to multiply the first number of one vector with the first number in the other. Then you do the same with the second and third numbers in each vector. Then you add up the three values and the answer is the dot product.
Example Question #141 : Regions
Find the dot product of and .
None of these
The way to find a dot product is to multiply the first number of one vector with the first number in the other. Then you do the same with the second and third numbers in each vector. Then you add up the three values and the answer is the dot product.
Example Question #63 : Area
Find the indefinite integral
Recall the power rule for integration:
Break apart the integral into three integrals:
Integrate term by term,
Now add the constant of integration to the expression.
Example Question #143 : Regions
Find the indefinite integral below
First, break apart the fraction into three fractions with each term having in its denominator.
Then cancel the variables where you can and integrate term by term. If you are left with any variables in the denominator then put them in the numerator with a negative sign.
You get:
Integrate these by the power rule,
We get: