Calculus Examples

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Step 1
Solve by substitution to find the intersection between the curves.
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Step 1.1
Eliminate the equal sides of each equation and combine.
Step 1.2
Solve for .
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Step 1.2.1
Move all terms containing to the left side of the equation.
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Step 1.2.1.1
Subtract from both sides of the equation.
Step 1.2.1.2
Combine the opposite terms in .
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Step 1.2.1.2.1
Subtract from .
Step 1.2.1.2.2
Add and .
Step 1.2.2
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
Step 1.2.3
Simplify .
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Step 1.2.3.1
Rewrite as .
Step 1.2.3.2
Pull terms out from under the radical, assuming positive real numbers.
Step 1.2.4
The complete solution is the result of both the positive and negative portions of the solution.
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Step 1.2.4.1
First, use the positive value of the to find the first solution.
Step 1.2.4.2
Next, use the negative value of the to find the second solution.
Step 1.2.4.3
The complete solution is the result of both the positive and negative portions of the solution.
Step 1.3
Evaluate when .
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Step 1.3.1
Substitute for .
Step 1.3.2
Substitute for in and solve for .
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Step 1.3.2.1
Remove parentheses.
Step 1.3.2.2
Remove parentheses.
Step 1.3.2.3
Add and .
Step 1.4
Evaluate when .
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Step 1.4.1
Substitute for .
Step 1.4.2
Substitute for in and solve for .
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Step 1.4.2.1
Remove parentheses.
Step 1.4.2.2
Remove parentheses.
Step 1.4.2.3
Add and .
Step 1.5
The solution to the system is the complete set of ordered pairs that are valid solutions.
Step 2
The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.
Step 3
Integrate to find the area between and .
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Step 3.1
Combine the integrals into a single integral.
Step 3.2
Apply the distributive property.
Step 3.3
Combine the opposite terms in .
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Step 3.3.1
Subtract from .
Step 3.3.2
Add and .
Step 3.4
Split the single integral into multiple integrals.
Step 3.5
Apply the constant rule.
Step 3.6
Since is constant with respect to , move out of the integral.
Step 3.7
By the Power Rule, the integral of with respect to is .
Step 3.8
Simplify the answer.
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Step 3.8.1
Combine and .
Step 3.8.2
Substitute and simplify.
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Step 3.8.2.1
Evaluate at and at .
Step 3.8.2.2
Evaluate at and at .
Step 3.8.2.3
Simplify.
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Step 3.8.2.3.1
Multiply by .
Step 3.8.2.3.2
Multiply by .
Step 3.8.2.3.3
Add and .
Step 3.8.2.3.4
Raise to the power of .
Step 3.8.2.3.5
Cancel the common factor of and .
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Step 3.8.2.3.5.1
Factor out of .
Step 3.8.2.3.5.2
Cancel the common factors.
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Step 3.8.2.3.5.2.1
Factor out of .
Step 3.8.2.3.5.2.2
Cancel the common factor.
Step 3.8.2.3.5.2.3
Rewrite the expression.
Step 3.8.2.3.5.2.4
Divide by .
Step 3.8.2.3.6
Raise to the power of .
Step 3.8.2.3.7
Cancel the common factor of and .
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Step 3.8.2.3.7.1
Factor out of .
Step 3.8.2.3.7.2
Cancel the common factors.
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Step 3.8.2.3.7.2.1
Factor out of .
Step 3.8.2.3.7.2.2
Cancel the common factor.
Step 3.8.2.3.7.2.3
Rewrite the expression.
Step 3.8.2.3.7.2.4
Divide by .
Step 3.8.2.3.8
Multiply by .
Step 3.8.2.3.9
Add and .
Step 3.8.2.3.10
Multiply by .
Step 3.8.2.3.11
Subtract from .
Step 4
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