Science:Math Exam Resources/Courses/MATH103/April 2014/Question 10 (b)/Solution 1

Following the suggestion in the question, we will integrate by parts. It is much easier to differentiate $f(x)$ than integrate it (we'd have to integrate the integral) and so we let

$u=f(x)=\int _{1}^{x}\sin(t^{2})\mathrm {d} t.$

The remaining part must be ${\textrm {d}}v$ and so therefore $\mathrm {d} v=1\mathrm {d} x$. This is fairly straightforward to integrate and so we get

$v=x.$

To differentiate $u$ we will use the fundamental theorem of calculus which states that

${\frac {\textrm {d}}{{\textrm {d}}x}}\int _{a}^{x}f(y){\textrm {d}}y=f(x).$

Applying this to our function yields

$\mathrm {d} u=\sin(x^{2})\mathrm {d} x.$

Using the information we have obtained with integration by parts formula,

$\int _{a}^{b}u{\textrm {d}}v=\left.uv\right|_{a}^{b}-\int _{a}^{b}v{\textrm {d}}u$

we get

${\begin{aligned}\int _{0}^{1}f(x)\mathrm {d} x&=\left[x\int _{1}^{x}\sin(t^{2})\mathrm {d} t\right]_{0}^{1}-\int _{0}^{1}x\sin(x^{2})\mathrm {d} x\\&=(1)\int _{1}^{1}\sin(t^{2})\mathrm {d} t-(0)\int _{1}^{0}\sin(t^{2})\mathrm {d} t-\int _{0}^{1}x\sin(x^{2})\mathrm {d} x\\&=-\int _{0}^{1}x\sin(x^{2})\mathrm {d} x,\end{aligned}}.$

The simplification comes from the fact that

$\int _{a}^{a}g(t)\mathrm {d} t=0$

for any function $g(t)$. In our particular case, $g(t)=\sin(t^{2})$ and $a=1$. The last integral can be evaluated by substitution. Let $y=x^{2}$ so that $\mathrm {d} y=2x\mathrm {d} x$. If we change variables, we have to be careful with the bounds. When $x=0$ then $y=0^{2}=0$ and when $x=1$ then $y=1^{2}=1$. Therefore the bounds don't change and the integral becomes

$-\int _{0}^{1}x\sin(x^{2})\mathrm {d} x=-{\frac {1}{2}}\int _{0}^{1}\sin y\mathrm {d} y=\left.{\frac {1}{2}}\cos y\right|_{0}^{1}={\frac {1}{2}}(\cos 1-1).$