This bit is good : ” —Often, math class is boring because it lacks creativity. John Updike, 20th century novelist and short story writer, had this to say about how seemingly mundane tasks can be made more creative:

An activity becomes creative when the doer cares about doing it right, or better. [John Updike]

There is plenty of scope for making math more creative. Start with the many connections with art, music and sports, and aim to do it better!—”

When one teaches Calculus to the higher secondary school level pupils , their motivation is ,generally , high.
The problem is to turn their vision from the idea of * just
doing well in the Final Exams.* to taking interest in the mathematical background of Calculus as an intellectual advance of all mankind.
Any one can pass * examinations* ; what distillate one takes beyond the examination is mathematical CULTURE !!

Regards

ahmed
Manali,
Himachal Pradesh,
INDIA

In example 4, you can stop at line 2 and treat the integral of the product of the terms as the “remainder” term. In general, any time the product of two columns is something easily integrable (not in my spell checker) you can stop and take the integral.

Nicely worked examples using this (tabular) method here: http://www.mccd.edu/faculty/powerd/M4b/M4b_L3_IBP.htm

This method also works for Int (e^x sin(x)) when you realize that the 3rd line is the same as the 1st.

The total distance covered is 60t km, t being in hours.
(At one time only)

When you come and go back, it is 120t km.

You spent 3t time (2t+t) covering a distance of 120t km. Hence, your average velocity is 120t km/3t h= 40 kmph.

This in a nut shell:

Plot of y=b^x is the arbitrary curve being studied, where b is any arbitrary base (other than zero or +-1), (x,0) is a linear length along the x axis and (0,y) is a nonlinear length along the y axis.

Therefore we get for our arbitrary plotted curve being studied:

y=b^x
b=(x root of(y)
x=logb(y)

From here you can find the base b of an unknown log (if a point y and its corresponding point x are known) or any point for y if x and b are known with the same being true for finding x.

This used in a “real life” example for the decay (exponential fall) of a capacitors stored energy into a resistance:

b=(e^-1)=(x root of(y) (the base used is that of exponential fall)
y=(e^-x) (y=magnitude, in percent)
x=loge^-1(y)=(-1)loge(y) (x=time constant “Tau”=(rC))

(where rC corresponds to seconds in time)

Plotting limits y=1max & x=5max (when practically considered, other wise x=infinity or some arbitrary number usually greater than 5)

To find the 50% magnitude or half the voltage of the capacitor:
y=0.5 (the 50% mark or half voltage)
x=loge(2)=(-1)loge(0.5)

This math stuff interpreted comes out to be 0.69314 * rC equals the point in time that the voltage is one half of its prior maximum.

Boring I know but this helped me out a bunch so I thought I would share this with a real life use example as well.