Question

Help! I need an answer for the integral lower limit -1 upper limit 1 x(x^2 +1)^5
I got an answer of 1/12 (x^2+1)^6 +C

Answer 1

You have the correct antiderivative, but the answer you have isn't the value of the definite integral \(\large\displaystyle \int_{-1}^1 x(x^2+1)^5\,dx\).
Using the work you did, we have that
\[\large \int_{-1}^1 x(x^2+1)^5\,dx = \left.\left[\frac{1}{12}(x^2+1)^6\right]\right|_{-1}^1\]
and now use the fundamental theorem of calculus to evaluate this. Does this make sense? :-)

Once you do this, I'll show you an even shorter way to do this problem. :-)

Answer 2

I'm not sure. I'm so confused right now! Can you show me?

Answer 3

the shorter way.

Answer 4

Thank you!

Answer 5

You do need to be familiar with the fundamental theorem of calculus anyways, so first note that
\[\large \begin{aligned} \left.\left[\frac{1}{12}(x^2+1)^6\right]\right|_{-1}^1 &= \left(\frac{1}{12}(1^2+1)^6\right) - \left(\frac{1}{12}((-1)^2+1)^6\right)\\ &= \frac{2^6}{12} - \frac{2^6}{12}\\ &= 0\end{aligned}\]

----------------

The shorter way is to make an observation about the function you're integrating. Take \(\large f(x) = x(x^2+1)^5\). Note that
\[\large f(-x) = -x((-x)^2+1)^5 = -x(x^2+1)^5=-f(x)\]
Thus the function we're integrating is ODD. Then, if we evaluate the definite integral of an odd function \(\large f(x)\) over a symmetric interval \(\large [-a,a]\), the result is ZERO, i.e.
\[\large \int_{-a}^a f(x)\,dx =0\]
Hence, in your problem, you're integrating the odd function \(\large f(x)=x(x^2+1)^5\) over the interval \(\large [-1,1]\). Therefore, you have
\[\large \int_{-1}^1x(x^2+1)^5\,dx = 0\]
by the property I mentioned above.

There is a similar property for even functions: If \(\large f(x)\) is even, then
\[\large \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx.\]
I hope this makes sense!