AI HL · AHL 3
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Further integration: substitution and boundary conditions
AHL 3.3
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\(\text{Find }\int\frac{x}{5x^2+2}\,dx\).
Have a go first. Then open the working one line at a time.
\(g(x)=5x^2+2,\ g'(x)=10x\)
The top is x. It is a multiple of g′, so the answer has ln.
\(u=5x^2+2,\ du=10x\,dx,\ dx=\frac{du}{10x}\)
Replace dx. The x on top cancels the x in dx.
\(\int\frac{x}{u}\cdot\frac{du}{10x}=\frac{1}{10}\int\frac{1}{u}\,du=\frac{1}{10}\ln|u|+c\)
The integral of 1/u is ln|u|.
\(\frac{1}{10}\ln|5x^2+2|+c\)
Write the answer in x. Here 5x² + 2 is always positive.
\(\text{Find }\int x^2\sqrt{2x^3+1}\,dx\).
Have a go first. Then open the working one line at a time.
\(u=2x^3+1,\ du=6x^2\,dx\)
The inside of the root is u. Its derivative has x², which is in the question.
\(x^2\,dx=\frac{1}{6}du,\quad \int\frac{1}{6}u^{1/2}\,du\)
The whole integral is now in u.
\(\frac{1}{6}\cdot\frac{u^{3/2}}{3/2}=\frac{1}{9}u^{3/2}+c\)
Add 1 to the power and divide by the new power. 6 × 3/2 = 9.
\(\frac{1}{9}(2x^3+1)^{3/2}+c\)
A cube root works the same way, with the power 1/3.
\(\text{Find }\int 10x^4e^{-4x^5}\,dx\).
Have a go first. Then open the working one line at a time.
\(u=-4x^5,\ du=-20x^4\,dx\)
The power of e is u. Its derivative is a multiple of x⁴.
\(10x^4\,dx=-\frac{1}{2}du,\quad \int-\frac{1}{2}e^u\,du\)
10 divided by −20 is −1/2.
\(-\frac{1}{2}e^{-4x^5}+c\)
The integral of \(e^u\) is \(e^u\). Put x back.
\(\frac{dy}{dx}=9\sqrt{x}\), and y = 50 when x = 4. Find y in terms of x.
Have a go first. Then open the working one line at a time.
\(y=\int 9x^{1/2}\,dx=9\cdot\frac{x^{3/2}}{3/2}+c=6x^{3/2}+c\)
Rewrite the root as a power, then add 1 to the power and divide.
\(50=6(4)^{3/2}+c=48+c\)
Put x = 4 and y = 50 into y, not into dy/dx.
\(c=2\)
Now every term of y is known.
\(y=6x^{3/2}+2\)
Write the full expression.
\(f'(x)=\frac{4}{x}\) and f(−2) = 6. Find the exact value of \(f(-6)\), then write it to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(f(x)=4\ln|x|+c\)
The values of x here are negative, so keep the absolute value.
\(f(-2)=4\ln 2+c=6,\ c=6-4\ln 2\)
ln|−2| = ln 2.
\(f(-6)=4\ln 6+6-4\ln 2=6+4\ln 3\)
Use ln 6 − ln 2 = ln 3.
From the calculator: 6 + 4 ln 3 = 10.4
The exact value is 6 + 4 ln 3.
\(f'(x)=\frac{6\ln x}{x}\) for x > 0, and f(e) = 10. Find f(x), then find \(f(e^2)\).
Have a go first. Then open the working one line at a time.
\(u=\ln x,\ du=\frac{1}{x}dx\)
The derivative of ln x is 1/x, and 1/x is already in the question.
\(\int 6u\,du=3u^2+c=3(\ln x)^2+c\)
Put ln x back.
\(f(e)=3(1)^2+c=10,\ c=7\)
ln e = 1.
\(f(e^2)=3(2)^2+7=19\)
ln e² = 2.