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Tom's question at Yahoo! Answers regarding solving for a limit of integration

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Tom,

Since I don't know the value the definite integral is to have, I will use $I$:

\(\displaystyle 5000\int_0^x 1-\frac{100}{(t+10)^2}\,dt=I\)

First, let's divide through by 5000:

\(\displaystyle \int_0^x 1-\frac{100}{(t+10)^2}\,dt=\frac{I}{5000}\)

Next, let's use the anti-derivative form of the FTOC on the left side:

\(\displaystyle \left[t+\frac{100}{t+10} \right]_0^x=\frac{I}{5000}\)

\(\displaystyle \left(x+\frac{100}{x+10} \right)-\left(0+\frac{100}{0+10} \right)=\frac{I}{5000}\)

\(\displaystyle x+\frac{100}{x+10}-10-\frac{I}{5000}=0\)

Now, multiply through by $x+10$:

\(\displaystyle x(x+10)+100-\left(10+\frac{I}{5000} \right)(x+10)=0\)

Arrange in standard quadratic form:

\(\displaystyle x^2+10x+100-10x-100-\frac{I}{5000}x-\frac{I}{500}=0\)

\(\displaystyle 5000x^2-Ix-10I=0\)

Applying the quadratic formula, we find:

\(\displaystyle x=\frac{I\pm\sqrt{I^2+200000I}}{10000}\)

Now, you just need to substitute the value of $I$ to find the two possible values of $x$, taking care not to cross the singularity in the original integrand.

In order for $x=10$, we find that we require \(\displaystyle I=25000\), however, this also allows $x=-5$.

To Tom and any other guests viewing this topic, I invite and encourage you to post other calculus questions here in our Calculus forum.

Best Regards,

Mark.