Calculus Examples

Evaluate the Summation sum from k=0 to 6 of -k^2-9
6k=0-k2-9
Step 1
Split the summation to make the starting value of k equal to 1.
6k=0-k2-9=6k=1-k2-9+0k=0-k2-9
Step 2
Evaluate 6k=1-k2-9.
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Step 2.1
Split the summation into smaller summations that fit the summation rules.
6k=1-k2-9=-16k=1k2+6k=1-9
Step 2.2
Evaluate 16k=1k2.
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Step 2.2.1
The formula for the summation of a polynomial with degree 2 is:
nk=1k2=n(n+1)(2n+1)6
Step 2.2.2
Substitute the values into the formula and make sure to multiply by the front term.
(-1)(6(6+1)(26+1)6)
Step 2.2.3
Simplify.
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Step 2.2.3.1
Simplify the numerator.
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Step 2.2.3.1.1
Add 6 and 1.
-167(26+1)6
Step 2.2.3.1.2
Combine exponents.
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Step 2.2.3.1.2.1
Multiply 6 by 7.
-142(26+1)6
Step 2.2.3.1.2.2
Multiply 2 by 6.
-142(12+1)6
-142(12+1)6
Step 2.2.3.1.3
Add 12 and 1.
-142136
-142136
Step 2.2.3.2
Simplify the expression.
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Step 2.2.3.2.1
Multiply 42 by 13.
-1(5466)
Step 2.2.3.2.2
Divide 546 by 6.
-191
Step 2.2.3.2.3
Multiply -1 by 91.
-91
-91
-91
-91
Step 2.3
Evaluate 6k=1-9.
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Step 2.3.1
The formula for the summation of a constant is:
nk=1c=cn
Step 2.3.2
Substitute the values into the formula.
(-9)(6)
Step 2.3.3
Multiply -9 by 6.
-54
-54
Step 2.4
Add the results of the summations.
-91-54
Step 2.5
Subtract 54 from -91.
-145
-145
Step 3
Evaluate 0k=0-k2-9.
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Step 3.1
Expand the series for each value of k.
-02-9
Step 3.2
Simplify.
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Step 3.2.1
Raise 0 to the power of 2.
-0-9
Step 3.2.2
Multiply -1 by 0.
0-9
Step 3.2.3
Subtract 9 from 0.
-9
-9
-9
Step 4
Replace the summations with the values found.
-145-9
Step 5
Subtract 9 from -145.
-154
 [x2  12  π  xdx ]