Dear Tom,
I would like to know if your program also solves integrals consisting only of symbols like x, a or b in both definite and indefinite matters?
Example:
or
Would solving does definite integrals be possible using your programs showing the solution process step by step?
Kind regards,
Ben
Answer
The first integral, sounds like your professor didn’t teach you, so you will want my programs to teach you or me to teach you. A common endeavor in my experience. When you see the da that’s “with respect to” in other words, you are integrating that integral with respect to “a”. In my programs you’d change all the “a’s” to x”s.
Now could you solve the integral if it looked like:
x x
∫ (1/x)dx or ∫ ( 1/[(c+b)-(k+x)] ) dx
o o
Now these would never appear on a test unless your professor demonstrated them on the black board, so they are either homework or experimenting off the internet. Both integrals are definite integrals because they show a range to integrate over. When you solve the integrals you have the area under that functions curve. That’s all integrals do is solve area under curves. You can graph 1/x on your calculator and you can see what the first integrals function looks like. Now, anytime you see a function without an exponent in the denominator, it is a natural log answer. The reason is is that you can’t integrate term with a division sign.
In this case the x has an exponent of 1 ( x^1), you always have to move the denominator to the numerator to eliminate the division sign before you can integrate. If you move the x^1 to the numerator to eliminate the division sign it becomes x^(-1) so when you do the integration process which is always add 1 to the exponent and divide the term by that answer, However when you add 1 to a -1 you get zero, and any term to the zero exponent = 1, so that’s why the natural log comes in. In the first integral above = ln(x), then you can enter the ranges for x that you are given, do the top range first and subtract the bottom range to get the answer (area under the curve of the function). The second integral above you’d break it into two integrals ∫[1/(c+b)] dx which equals x/(c+b) and then the second term 1/[(k+x)]dx which equals ln(k+x).
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