Calculus Examples

Determine Convergence with the Integral Test
k=1kek2k=1kek2
Step 1
Check if the function is continuous over the summation bounds.
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Step 1.1
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Interval Notation:
(-,)(,)
Set-Builder Notation:
{k|k}
Step 1.2
f(k) is continuous on [1,).
The function is continuous.
The function is continuous.
Step 2
Check if the function is positive over the bounds.
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Step 2.1
Set up an inequality.
kek2>0
Step 2.2
Solve the inequality.
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Step 2.2.1
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
k=0
ek2=0
Step 2.2.2
Set k equal to 0.
k=0
Step 2.2.3
Set ek2 equal to 0 and solve for k.
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Step 2.2.3.1
Set ek2 equal to 0.
ek2=0
Step 2.2.3.2
Solve ek2=0 for k.
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Step 2.2.3.2.1
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(ek2)=ln(0)
Step 2.2.3.2.2
The equation cannot be solved because ln(0) is undefined.
Undefined
Step 2.2.3.2.3
There is no solution for ek2=0
No solution
No solution
No solution
Step 2.2.4
The final solution is all the values that make kek2>0 true.
k=0
Step 2.2.5
The solution consists of all of the true intervals.
k>0
k>0
k>0
Step 3
Determine where the function is decreasing.
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Step 3.1
Write kek2 as a function.
f(k)=kek2
Step 3.2
Find the first derivative.
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Step 3.2.1
Find the first derivative.
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Step 3.2.1.1
Differentiate using the Product Rule which states that ddk[f(k)g(k)] is f(k)ddk[g(k)]+g(k)ddk[f(k)] where f(k)=k and g(k)=ek2.
kddk[ek2]+ek2ddk[k]
Step 3.2.1.2
Differentiate using the chain rule, which states that ddk[f(g(k))] is f(g(k))g(k) where f(k)=ek and g(k)=k2.
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Step 3.2.1.2.1
To apply the Chain Rule, set u as k2.
k(ddu[eu]ddk[k2])+ek2ddk[k]
Step 3.2.1.2.2
Differentiate using the Exponential Rule which states that ddu[au] is auln(a) where a=e.
k(euddk[k2])+ek2ddk[k]
Step 3.2.1.2.3
Replace all occurrences of u with k2.
k(ek2ddk[k2])+ek2ddk[k]
k(ek2ddk[k2])+ek2ddk[k]
Step 3.2.1.3
Differentiate using the Power Rule which states that ddk[kn] is nkn-1 where n=2.
k(ek2(2k))+ek2ddk[k]
Step 3.2.1.4
Raise k to the power of 1.
k1k(ek2(2))+ek2ddk[k]
Step 3.2.1.5
Raise k to the power of 1.
k1k1(ek2(2))+ek2ddk[k]
Step 3.2.1.6
Use the power rule aman=am+n to combine exponents.
k1+1(ek2(2))+ek2ddk[k]
Step 3.2.1.7
Simplify the expression.
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Step 3.2.1.7.1
Add 1 and 1.
k2(ek2(2))+ek2ddk[k]
Step 3.2.1.7.2
Move 2 to the left of ek2.
k2(2ek2)+ek2ddk[k]
k2(2ek2)+ek2ddk[k]
Step 3.2.1.8
Differentiate using the Power Rule which states that ddk[kn] is nkn-1 where n=1.
k2(2ek2)+ek21
Step 3.2.1.9
Multiply ek2 by 1.
k2(2ek2)+ek2
Step 3.2.1.10
Simplify.
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Step 3.2.1.10.1
Reorder terms.
2ek2k2+ek2
Step 3.2.1.10.2
Reorder factors in 2ek2k2+ek2.
f(k)=2k2ek2+ek2
f(k)=2k2ek2+ek2
f(k)=2k2ek2+ek2
Step 3.2.2
The first derivative of f(k) with respect to k is 2k2ek2+ek2.
2k2ek2+ek2
2k2ek2+ek2
Step 3.3
Set the first derivative equal to 0 then solve the equation 2k2ek2+ek2=0.
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Step 3.3.1
Set the first derivative equal to 0.
2k2ek2+ek2=0
Step 3.3.2
Factor ek2 out of 2k2ek2+ek2.
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Step 3.3.2.1
Factor ek2 out of 2k2ek2.
ek2(2k2)+ek2=0
Step 3.3.2.2
Multiply by 1.
ek2(2k2)+ek21=0
Step 3.3.2.3
Factor ek2 out of ek2(2k2)+ek21.
ek2(2k2+1)=0
ek2(2k2+1)=0
Step 3.3.3
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
ek2=0
2k2+1=0
Step 3.3.4
Set ek2 equal to 0 and solve for k.
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Step 3.3.4.1
Set ek2 equal to 0.
ek2=0
Step 3.3.4.2
Solve ek2=0 for k.
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Step 3.3.4.2.1
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(ek2)=ln(0)
Step 3.3.4.2.2
The equation cannot be solved because ln(0) is undefined.
Undefined
Step 3.3.4.2.3
There is no solution for ek2=0
No solution
No solution
No solution
Step 3.3.5
Set 2k2+1 equal to 0 and solve for k.
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Step 3.3.5.1
Set 2k2+1 equal to 0.
2k2+1=0
Step 3.3.5.2
Solve 2k2+1=0 for k.
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Step 3.3.5.2.1
Subtract 1 from both sides of the equation.
2k2=-1
Step 3.3.5.2.2
Divide each term in 2k2=-1 by 2 and simplify.
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Step 3.3.5.2.2.1
Divide each term in 2k2=-1 by 2.
2k22=-12
Step 3.3.5.2.2.2
Simplify the left side.
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Step 3.3.5.2.2.2.1
Cancel the common factor of 2.
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Step 3.3.5.2.2.2.1.1
Cancel the common factor.
2k22=-12
Step 3.3.5.2.2.2.1.2
Divide k2 by 1.
k2=-12
k2=-12
k2=-12
Step 3.3.5.2.2.3
Simplify the right side.
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Step 3.3.5.2.2.3.1
Move the negative in front of the fraction.
k2=-12
k2=-12
k2=-12
Step 3.3.5.2.3
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
k=±-12
Step 3.3.5.2.4
Simplify ±-12.
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Step 3.3.5.2.4.1
Rewrite -12 as i2122.
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Step 3.3.5.2.4.1.1
Rewrite -1 as i2.
k=±i212
Step 3.3.5.2.4.1.2
Rewrite 1 as 12.
k=±i2122
k=±i2122
Step 3.3.5.2.4.2
Pull terms out from under the radical.
k=±i122
Step 3.3.5.2.4.3
One to any power is one.
k=±i12
Step 3.3.5.2.4.4
Rewrite 12 as 12.
k=±i12
Step 3.3.5.2.4.5
Any root of 1 is 1.
k=±i12
Step 3.3.5.2.4.6
Multiply 12 by 22.
k=±i(1222)
Step 3.3.5.2.4.7
Combine and simplify the denominator.
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Step 3.3.5.2.4.7.1
Multiply 12 by 22.
k=±i222
Step 3.3.5.2.4.7.2
Raise 2 to the power of 1.
k=±i2212
Step 3.3.5.2.4.7.3
Raise 2 to the power of 1.
k=±i22121
Step 3.3.5.2.4.7.4
Use the power rule aman=am+n to combine exponents.
k=±i221+1
Step 3.3.5.2.4.7.5
Add 1 and 1.
k=±i222
Step 3.3.5.2.4.7.6
Rewrite 22 as 2.
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Step 3.3.5.2.4.7.6.1
Use nax=axn to rewrite 2 as 212.
k=±i2(212)2
Step 3.3.5.2.4.7.6.2
Apply the power rule and multiply exponents, (am)n=amn.
k=±i22122
Step 3.3.5.2.4.7.6.3
Combine 12 and 2.
k=±i2222
Step 3.3.5.2.4.7.6.4
Cancel the common factor of 2.
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Step 3.3.5.2.4.7.6.4.1
Cancel the common factor.
k=±i2222
Step 3.3.5.2.4.7.6.4.2
Rewrite the expression.
k=±i221
k=±i221
Step 3.3.5.2.4.7.6.5
Evaluate the exponent.
k=±i22
k=±i22
k=±i22
Step 3.3.5.2.4.8
Combine i and 22.
k=±i22
k=±i22
Step 3.3.5.2.5
The complete solution is the result of both the positive and negative portions of the solution.
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Step 3.3.5.2.5.1
First, use the positive value of the ± to find the first solution.
k=i22
Step 3.3.5.2.5.2
Next, use the negative value of the ± to find the second solution.
k=-i22
Step 3.3.5.2.5.3
The complete solution is the result of both the positive and negative portions of the solution.
k=i22,-i22
k=i22,-i22
k=i22,-i22
k=i22,-i22
Step 3.3.6
The final solution is all the values that make ek2(2k2+1)=0 true.
k=i22,-i22
k=i22,-i22
Step 3.4
There are no values of k in the domain of the original problem where the derivative is 0 or undefined.
No critical points found
Step 3.5
No points make the derivative f(k)=2k2ek2+ek2 equal to 0 or undefined. The interval to check if f(k)=kek2 is increasing or decreasing is (-,).
(-,)
Step 3.6
Substitute any number, such as 1, from the interval (-,) in the derivative f(k)=2k2ek2+ek2 to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval (-,). If the result is positive, the graph is increasing on the interval (-,).
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Step 3.6.1
Replace the variable k with 1 in the expression.
f(1)=2(1)2e(1)2+e(1)2
Step 3.6.2
Simplify the result.
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Step 3.6.2.1
Simplify each term.
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Step 3.6.2.1.1
One to any power is one.
f(1)=2(1e(1)2)+e(1)2
Step 3.6.2.1.2
Multiply 2 by 1.
f(1)=2e(1)2+e(1)2
Step 3.6.2.1.3
One to any power is one.
f(1)=2e+e(1)2
Step 3.6.2.1.4
Simplify.
f(1)=2e+e(1)2
Step 3.6.2.1.5
One to any power is one.
f(1)=2e+e
Step 3.6.2.1.6
Simplify.
f(1)=2e+e
f(1)=2e+e
Step 3.6.2.2
Add 2e and e.
f(1)=3e
Step 3.6.2.3
The final answer is 3e.
3e
3e
3e
Step 3.7
The result of substituting 1 into f(k)=2k2ek2+ek2 is 3e, which is positive, so the graph is increasing on the interval (-,).
Increasing on (-,) since 2k2ek2+ek2>0
Step 3.8
Increasing over the interval (-,) means that the function is always increasing.
Always Increasing
Always Increasing
Step 4
The integral test does not apply because the function is not always decreasing from 1 to .
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 [x2  12  π  xdx ] 
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