8.3 Finding the constant of integration
Integration alone gives a whole family of curves, one for each value of \(c\), all with the same shape but sitting at different heights. One extra piece of information — a single point the curve passes through — picks out which member of the family is meant.
Example 8.10. A curve \(y=f(x)\) passes through the point \((9,10)\), and \[f'(x)=\frac {x-3}{\sqrt {x}}.\] Find the equation of the curve.
Solution. Step 1 — rewrite in a form that can be integrated. Divide each term of the numerator by \(x^{\frac {1}{2}}\): \[f'(x)=\frac {x}{x^{\frac {1}{2}}}-\frac {3}{x^{\frac {1}{2}}} =x^{\frac {1}{2}}-3x^{-\frac {1}{2}}.\]
Step 2 — integrate. \[f(x)=\frac {x^{\frac {3}{2}}}{\frac {3}{2}}-\frac {3x^{\frac {1}{2}}}{\frac {1}{2}}+c =\frac {2}{3}x^{\frac {3}{2}}-6\sqrt {x}+c.\]
Step 3 — use the point to find \(c\). The curve passes through \((9,10)\), so substituting \(x=9\) must give \(y=10\). Note \(\sqrt {9}=3\), so \(9^{\frac {3}{2}}=3^3=27\): \[10=\frac {2}{3}(27)-6(3)+c=18-18+c=c.\] So \(c=10\).
Step 4 — state the curve. \[\therefore \quad y=\frac {2}{3}x^{\frac {3}{2}}-6\sqrt {x}+10.\]
Note 8.11. The point is substituted after integrating, never before. Substituting into \(f'(x)\) would give the gradient at that point, which is a different question entirely.
8.3.1 Definite integrals
When an integral is given two limits, the constant of integration ceases to matter and the answer is a number rather than a function.
Definition 8.12. If \(F(x)\) is an integral of \(f(x)\), then the definite integral between \(x=a\) and \(x=b\) is \[\int _a^b f(x)\,dx=\Big [F(x)\Big ]_a^b=F(b)-F(a),\] where \(b\) is the upper limit and \(a\) the lower limit.
Note 8.13. No \(+c\) is written. If it were carried through, the subtraction would give \(\left (F(b)+c\right )-\left (F(a)+c\right )\), and the two copies of \(c\) would cancel. It is not that the constant is forgotten — it genuinely makes no difference here.
Example 8.14. Evaluate (a) \(\displaystyle \int _1^3\left (3x^2-4x+1\right )dx\), (b) \(\displaystyle \int _1^4\frac {2}{\sqrt {x}}\,dx\), (c) \(\displaystyle \int _0^{\frac {\pi }{2}}\cos x\,dx\).
Solution. (a) Integrate first, keeping the limits attached to the square bracket: \[\int _1^3\left (3x^2-4x+1\right )dx=\Big [x^3-2x^2+x\Big ]_1^3.\] Now substitute the upper limit, then the lower, and subtract: \[=\left (3^3-2(3)^2+3\right )-\left (1^3-2(1)^2+1\right )\] \[=\left (27-18+3\right )-\left (1-2+1\right )=12-0=12.\]
(b) Convert to a power first: \[\int _1^4 2x^{-\frac {1}{2}}\,dx=\left [\frac {2x^{\frac {1}{2}}}{\frac {1}{2}}\right ]_1^4 =\Big [4\sqrt {x}\Big ]_1^4=4(2)-4(1)=4.\]
(c) \[\int _0^{\frac {\pi }{2}}\cos x\,dx=\Big [\sin x\Big ]_0^{\frac {\pi }{2}} =\sin \frac {\pi }{2}-\sin 0=1-0=1.\] The limits are in radians, as they must be whenever calculus is done with trigonometric functions.
Note 8.15. The commonest error here is subtracting in the wrong order. It is always \(F(\text {upper})-F(\text {lower})\); reversing it changes the sign of the whole answer. Brackets around each substituted value, as in (a), prevent the second mistake — losing a sign when the lower value is itself negative.
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