Definite Integral

Show that the definite integral $$\underset{a}{\overset{b}{\int }}f(x)dx$$ for $$f(x)=sin(x)$$ on $$[\frac{\pi }{2},\frac{3\pi }{2}]$$ is 0.

syms x
int(sin(x),pi/2,3*pi/2)

ans = $$0$$

Definite Integrals in Maxima and Minima

To maximize $$F(a)={\int }_{-a}^a\sin (ax)\sin (x/a)dx$$ for $$a\ge 0$$, first, define the symbolic variables and assume that $$a\ge 0$$:

syms a x
assume(a >= 0);

Then, define the function to maximize:

F = int(sin(a*x)*sin(x/a),x,-a,a)

F = 
$$\{\begin{array}{cl}1-\frac{\sin \left(2\right)}{2}& \text{ if }a=1\\ \frac{2\hspace{0.17em}a\hspace{0.17em}\left(\sin \left(a^2\right)\hspace{0.17em}\cos \left(1\right)-a^2\hspace{0.17em}\cos \left(a^2\right)\hspace{0.17em}\sin \left(1\right)\right)}{a^4-1}& \text{ if }a\ne 1\end{array}$$

Note the special case here for $$a=1$$. To make computations easier, use assumeAlso to ignore this possibility (and later check that $$a=1$$ is not the maximum):

assumeAlso(a ~= 1);
F = int(sin(a*x)*sin(x/a),x,-a,a)

F = 
$$\frac{2\hspace{0.17em}a\hspace{0.17em}\left(\sin \left(a^2\right)\hspace{0.17em}\cos \left(1\right)-a^2\hspace{0.17em}\cos \left(a^2\right)\hspace{0.17em}\sin \left(1\right)\right)}{a^4-1}$$

Create a plot of $$F$$ to check its shape:

fplot(F,[0 10])

Use diff to find the derivative of $$F$$ with respect to $$a$$:

Fa = diff(F,a)

Fa = 
$$\begin{array}{l}\frac{2\hspace{0.17em}{\sigma }_1}{a^4-1}+\frac{2\hspace{0.17em}a\hspace{0.17em}\left(2\hspace{0.17em}a\hspace{0.17em}\cos \left(a^2\right)\hspace{0.17em}\cos \left(1\right)-2\hspace{0.17em}a\hspace{0.17em}\cos \left(a^2\right)\hspace{0.17em}\sin \left(1\right)+2\hspace{0.17em}{a}^3\hspace{0.17em}\sin \left(a^2\right)\hspace{0.17em}\sin \left(1\right)\right)}{a^4-1}-\frac{8\hspace{0.17em}{a}^4\hspace{0.17em}{\sigma }_1}{{\left(a^4-1\right)}^2}\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_1=\sin \left(a^2\right)\hspace{0.17em}\cos \left(1\right)-a^2\hspace{0.17em}\cos \left(a^2\right)\hspace{0.17em}\sin \left(1\right)\end{array}$$

The zeros of $$Fa$$ are the local extrema of $$F$$:

hold on
fplot(Fa,[0 10])
grid on

The maximum is between 1 and 2. Use vpasolve to find an approximation of the zero of $$Fa$$ in this interval:

a_max = vpasolve(Fa,a,[1,2])

a_max = $$1.5782881585233198075558845180583$$

Use subs to get the maximal value of the integral:

F_max = subs(F,a,a_max)

F_max = $$0.36730152527504169588661811770092\hspace{0.17em}\cos \left(1\right)+1.2020566879911789986062956284113\hspace{0.17em}\sin \left(1\right)$$

The result still contains exact numbers $$\sin (1)$$ and $$\cos (1)$$. Use vpa to replace these by numerical approximations:

vpa(F_max)

ans = $$1.2099496860938456039155811226054$$

Check that the excluded case $$a=1$$ does not result in a larger value:

vpa(int(sin(x)*sin(x),x,-1,1))

ans = $$0.54535128658715915230199006704413$$

Multiple Integration

Numerical integration over higher dimensional areas has special functions:

integral2(@(x,y) x.^2-y.^2,0,1,0,1)

ans = 4.0127e-19

There are no such special functions for higher-dimensional symbolic integration. Use nested one-dimensional integrals instead:

syms x y
int(int(x^2-y^2,y,0,1),x,0,1)

ans = $$0$$

Line Integrals

Define a vector field F in 3D space:

syms x y z
F(x,y,z) = [x^2*y*z, x*y, 2*y*z];

Next, define a curve:

syms t
ux(t) = sin(t);
uy(t) = t^2-t;
uz(t) = t;

The line integral of F along the curve u is defined as $$\int f\cdot du=\int f(ux(t),uy(t),uz(t))\cdot \frac{du}{dt}dt$$, where the $$\cdot$$ on the right-hand-side denotes a scalar product.

Use this definition to compute the line integral for $$t$$ from $$[0,1]$$

F_int = int(F(ux,uy,uz)*diff([ux;uy;uz],t),t,0,1)

F_int = 
$$\frac{19\hspace{0.17em}\cos \left(1\right)}{4}-\frac{\cos \left(3\right)}{108}-12\hspace{0.17em}\sin \left(1\right)+\frac{\sin \left(3\right)}{27}+\frac{395}{54}$$

Get a numerical approximation of this exact result:

vpa(F_int)

ans = $$-0.20200778585035447453044423341349$$