Find The Exact Value of A Definite Integral

[anemone]

Find the exact value of the following definite integral:

\(\displaystyle \int_0^2 (3x^2-3x+1)\cos (x^3-3x^2+4x-2)\,dx\)

 

[MarkFL]

Interesting problem! I noticed neither my calculator (TI-89) nor W|A could come up with the exact result. Here is my approach:

Spoiler

Let:

\(\displaystyle I=\int_0^2\left(3x^2-3x+1 \right)\cos\left(x^3-3x^2+4x-2 \right)\,dx\)

Observe that:

\(\displaystyle x^3-3x^2+4x-2=(x-1)^3+(x-1)\)

\(\displaystyle 3x^2-3x+1=3x(x-1)+1\)

So, let:

\(\displaystyle u=x-1\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+3u+1 \right)\cos\left(u^3+u \right)\,du\)

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+1 \right)\cos\left(u^3+u \right)\,du+3\int_{-1}^{1}u\cos\left(u^3+u \right)\,du\)

By the odd-function rule, this reduces to:

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+1 \right)\cos\left(u^3+u \right)\,du\)

Let:

\(\displaystyle v=u^3+u\,\therefore\,dv=\left(3u^2+1 \right)\,du\)

and we have (using the even-function rule):

\(\displaystyle I=2\int_0^2\cos(v)\,dv=2\sin(2)\approx1.8185948536513634\)

 

[anemone]

What a great solution, MarkFL!(Sun)

 

[Prove It]

Find the exact value of the following definite integral:

\(\displaystyle \int_0^2 (3x^2-3x+1)\cos (x^3-3x^2+4x-2)\,dx\)

I have a slightly different approach to Mark's...

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x^2 - 3x + 1 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \,dx } \\ = \int_0^2{ \left( 3x^2 - 6x + 4 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx} + \int_0^2{ \left( 3x - 3 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \,dx } \end{align*}\)

The first integral can be solved using \(\displaystyle \displaystyle \begin{align*} u = x^3 - 3x^2 + 4x - 2 \implies du = \left( 3x^2 - 6x + 4 \right) \, dx \end{align*}\) giving

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x^2 - 6x + 4 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx} &= \int_{-2}^2{ \cos{(u)}\,du } \\ &= \left[ \sin{(u)} \right]_{-2}^2 \\ &= \sin{(2)} - \sin{(-2)} \\ &= \sin{(2)} + \sin{(2)} \\ &= 2\sin{(2)} \end{align*}\)

As for the second

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x - 3 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx } &= 3 \int_0^2{ \left( x - 1 \right) \cos{ \left[ \left( x - 1 \right) \left( x^2 - 2x + 2 \right) \right] } \, dx } \\ &= 3\int_0^2{ \left( x - 1 \right) \cos{ \left\{ \left( x - 1 \right) \left[ \left( x - 1 \right) ^2 + 1 \right] \right\} } \, dx } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} v = x - 1 \implies dv = dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} 3\int_{-1}^1{ v \cos{ \left[ v \left( v^2 + 1 \right) \right] } \, dv } &= 0 \end{align*}\)

because it's an odd function integrated over an even region.Therefore, the original integral equals \(\displaystyle \displaystyle \begin{align*} 2\sin{(2)} \end{align*}\)

 

[CellCoree]

To find the exact value of this definite integral, we can use the fundamental theorem of calculus. First, we can integrate the given function (3x^2-3x+1)\cos (x^3-3x^2+4x-2) with respect to x, which will give us a new function. Then, we can evaluate this new function at the upper and lower limits of integration (0 and 2 in this case), and subtract the two values to get the exact value of the definite integral.

This process may require some algebraic manipulation and possibly integration by parts, depending on the complexity of the function. It is important to note that the exact value of a definite integral cannot always be found using analytical methods, in which case numerical methods such as numerical integration or approximation techniques may be used.

In summary, to find the exact value of a definite integral, we must integrate the given function, evaluate it at the upper and lower limits of integration, and subtract the two values. If the function is too complex to be evaluated analytically, numerical methods may be used instead.