The product rule, integrated, becomes integration by parts: trade $\int u\,dv$ for $uv-\int v\,du$. Choose $u$ well (ILATE) and the new integral is simpler; it also gives $\int e^x[f(x)+f'(x)]\,dx$ and the integrals of $\sqrt{a^2-x^2}$ and $\sqrt{x^2\pm a^2}$.
A medicine is absorbed so that the rate at which it enters the blood is $50te^{-0.5t}$ milligrams per hour: it rises, peaks and dies away. The total absorbed needs $\displaystyle\int te^{-0.5t}\,dt$. A mass on a spring that is being pushed in time with its own swing moves with velocity $t\cos t$; its position needs $\displaystyle\int t\cos t\,dt$. Both integrands are products of two different kinds of function, and neither is the derivative of anything obvious. Substitution does not help: there is no inner function whose derivative is lying around.
What does help is the product rule, read backwards. Differentiating a product $uv$ gives two terms; integrating the rule lets us swap an integral we cannot do, $\int u\,dv$, for one we hope we can, $\int v\,du$. The whole skill is in choosing which factor to call $u$, and this section gives a reliable rule for that — the ILATE order — together with three families of integrals that this method settles once and for all.
The Product Rule in Reverse
📐 Theorem 7.8 — Integration by parts
Let $u$ be differentiable with a continuous derivative on an interval $I$, and let $v$ be continuous on $I$. Then on $I$
$$\int u\,v\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx$$
In words: the integral of a product is first function × integral of the second, minus the integral of (derivative of the first) × (integral of the second). In the shorthand $dv=v\,dx$ this is
$$\int u\,dv=uv-\int v\,du$$
Why. Let $w$ be any antiderivative of $v$, so $w'=v$. By the product rule, $\dfrac{d}{dx}(uw)=u'w+uw'=u'w+uv$. So $uv=\dfrac{d}{dx}(uw)-u'w$, and integrating both sides,
$$\int uv\,dx=uw-\int u'w\,dx$$
which is the theorem with $w=\int v\,dx$. No constant is needed when you write $\int v\,dx$ in the middle of the formula: any constant there cancels out (try it), so add a single $+\,C$ at the very end.
🧭 Problem-solving strategy — choosing the first function (ILATE)
Call $u$ (the first function) the factor that comes first in this list, and integrate the other:
I — Inverse trigonometric$\sin^{-1}x$, $\tan^{-1}x$, …
L — Logarithmic$\log x$, $\log(1+x^2)$, …
A — Algebraic$x$, $x^2$, $\sqrt x$, polynomials
T — Trigonometric$\sin x$, $\cos2x$, $\sec^2x$, …
E — Exponential$e^x$, $e^{-2x}$, $2^x$
Why it works: functions near the top get simpler when differentiated (the derivative of $\log x$ or $\tan^{-1}x$ is algebraic; a power of $x$ drops by one), while functions near the bottom are easy to integrate and do not get worse. If the first attempt makes the new integral harder, swap the roles. Some books use the order LIATE; for every integral in this course the two orders give the same choice.
✏️ Example 1 — A good choice and a bad one
Find $\displaystyle\int x\cos x\,dx$. Then show what happens with the opposite choice of first function.
💡 Show solution▶
ILATE $x$ is Algebraic and $\cos x$ is Trigonometric; A comes before T, so $u=x$ and the second function is $\cos x$, with $\int\cos x\,dx=\sin x$.
Parts $$\int x\cos x\,dx=x\sin x-\int1\cdot\sin x\,dx=x\sin x+\cos x+C$$
Bad choice With $u=\cos x$ and the second function $x$ (integral $\tfrac{x^2}{2}$):
$$\int x\cos x\,dx=\frac{x^2}{2}\cos x+\int\frac{x^2}{2}\sin x\,dx$$
The new integral has $x^2$ instead of $x$ — harder than where we started. The formula is still true; it is just useless. Try both choices in Interactive 1.
✏️ Example 2 — Parts twice
Find $\displaystyle\int x^2e^x\,dx$.
💡 Show solution▶
First $u=x^2$ (A before E), second function $e^x$: $\displaystyle\int x^2e^x\,dx=x^2e^x-\int2xe^x\,dx$. The power of $x$ has dropped from $2$ to $1$.
Second Again with $u=x$: $\displaystyle\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x$.
(a) $u=\log x$, second function $1$ (integral $x$), on $x\gt0$:
$$\int\log x\,dx=x\log x-\int\frac1x\cdot x\,dx=x\log x-x+C$$
(b) $u=\tan^{-1}x$: $\displaystyle\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx$, and the last integral is $\tfrac12\log(1+x^2)$ by the $\tfrac{f'}{f}$ rule. So
$$\int\tan^{-1}x\,dx=x\tan^{-1}x-\tfrac12\log\left(1+x^2\right)+C$$
(c) $u=\sin^{-1}x$: $\displaystyle\int\sin^{-1}x\,dx=x\sin^{-1}x-\int\frac{x}{\sqrt{1-x^2}}\,dx$. With $w=1-x^2$, $\displaystyle\int\frac{x\,dx}{\sqrt{1-x^2}}=-\sqrt{1-x^2}$. So
$$\int\sin^{-1}x\,dx=x\sin^{-1}x+\sqrt{1-x^2}+C,\qquad -1\lt x\lt1$$
Interactive 1 — Integration-by-Parts Helper: Choose the First Function
Pick an integral and choose which factor is the first function $u$. The helper shows each factor's ILATE letter, writes out $u$, $du$, $v$ and the formula $uv-\int v\,du$, and judges whether the new integral is simpler. The graph is the check by differentiation: the black integrand and the dashed red derivative of the final answer must coincide. With the bad choice, the blue curve is the new integrand $v\,\dfrac{du}{dx}$ that you would be left with. Play traces the derivative check.
ILATE letters—
$u$, $du$, $v$—
after one step—
Two Special Patterns: $e^x[f(x)+f'(x)]$ and Integrals That Come Back
Look at $\displaystyle\int e^x(\sin x+\cos x)\,dx$. Integrate $\int e^x\sin x\,dx$ by parts with $u=\sin x$: you get $e^x\sin x-\int e^x\cos x\,dx$, and the leftover integral cancels the other half of the integrand exactly. This happens whenever the bracket is a function plus its own derivative.
📐 Theorem 7.9 — The pattern $e^x[f(x)+f'(x)]$
If $f$ is differentiable with a continuous derivative on an interval $I$, then on $I$
$$\int e^x\left[f(x)+f'(x)\right]dx=e^xf(x)+C$$
Why. By the product rule, $\dfrac{d}{dx}\left(e^xf(x)\right)=e^xf(x)+e^xf'(x)$ $=e^x\left[f(x)+f'(x)\right]$. (Equivalently: integrate $\int e^xf(x)\,dx$ by parts with $u=f(x)$; the integral $\int e^xf'(x)\,dx$ that appears cancels.) The skill is to see the bracket as $f+f'$, which sometimes needs a little algebra first.
(b) $\dfrac{x-1}{x^2}=\dfrac1x-\dfrac1{x^2}$, and $\left(\tfrac1x\right)'=-\tfrac1{x^2}$. So $f=\tfrac1x$ and the answer is $\dfrac{e^x}{x}+C$.
(c) $\dfrac{1+x\log x}{x}=\log x+\dfrac1x$, with $f=\log x$, $f'=\tfrac1x$: $\ e^x\log x+C$ (on $x\gt0$).
(d) Write $x=(1+x)-1$: $\dfrac{x}{(1+x)^2}=\dfrac{1}{1+x}-\dfrac{1}{(1+x)^2}$. With $f=\dfrac1{1+x}$, $f'=-\dfrac1{(1+x)^2}$, the answer is $\dfrac{e^x}{1+x}+C$.
A different thing happens with $\displaystyle\int e^x\sin x\,dx$ on its own: integrating by parts twice brings back the integral you started with. Rather than going round in circles, call it $I$ and solve for it.
✏️ Example 5 — A cyclic integral
Find $I=\displaystyle\int e^x\sin x\,dx$.
💡 Show solution▶
Once $u=\sin x$, second function $e^x$: $\ I=e^x\sin x-\displaystyle\int e^x\cos x\,dx$.
Twice In the new integral take $u=\cos x$ — keep the same kind of first function, or you will undo the first step:
$$\int e^x\cos x\,dx=e^x\cos x+\int e^x\sin x\,dx=e^x\cos x+I$$
Solve $I=e^x\sin x-e^x\cos x-I$, so $2I=e^x(\sin x-\cos x)$ and
$$\int e^x\sin x\,dx=\frac12e^x(\sin x-\cos x)+C$$
Check The derivative of $\tfrac12e^x(\sin x-\cos x)$ is $\tfrac12e^x(\sin x-\cos x)$ $+\tfrac12e^x(\cos x+\sin x)$ $=e^x\sin x$ ✓.
The same two-step method gives the general results, worth knowing for JEE: for $a^2+b^2\ne0$,
$$\int e^{ax}\sin bx\,dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C$$
$$\int e^{ax}\cos bx\,dx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C$$
Integrals of $\sqrt{a^2-x^2}$ and $\sqrt{x^2\pm a^2}$
Section 7.2 integrated one over these square roots. The square roots themselves are integrated by parts, using the cyclic trick.
📐 Theorem 7.10 — Three square-root integrals
For $a\gt0$, on any interval where the root is defined:
root of a² − x²$\displaystyle\int\sqrt{a^2-x^2}\,dx$ $\displaystyle=\frac x2\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$
root of x² + a²$\displaystyle\int\sqrt{x^2+a^2}\,dx$ $\displaystyle=\frac x2\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C$
root of x² − a²$\displaystyle\int\sqrt{x^2-a^2}\,dx$ $\displaystyle=\frac x2\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C$
Why (the middle one). Let $I=\displaystyle\int\sqrt{x^2+a^2}\,dx$. Take $u=\sqrt{x^2+a^2}$ and second function $1$; then $\dfrac{du}{dx}=\dfrac{x}{\sqrt{x^2+a^2}}$ and
$$I=x\sqrt{x^2+a^2}-\int\frac{x^2}{\sqrt{x^2+a^2}}\,dx$$
Write $x^2=(x^2+a^2)-a^2$ in the last integrand: it becomes $\sqrt{x^2+a^2}-\dfrac{a^2}{\sqrt{x^2+a^2}}$. So
$$I=x\sqrt{x^2+a^2}-I+a^2\log\left|x+\sqrt{x^2+a^2}\right|$$
using special integral (6) of Section 7.2. Solving $2I=\cdots$ gives the formula. The other two are proved the same way, with special integrals (4) and (5).
(a) $x^2+2x+5=(x+1)^2+2^2$. With $t=x+1$, $a=2$ in the middle formula ($\tfrac{a^2}{2}=2$):
$$\frac{x+1}{2}\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C$$
(b) $3-2x-x^2=4-(x+1)^2$. With $t=x+1$, $a=2$ in the first formula:
$$\frac{x+1}{2}\sqrt{3-2x-x^2}+2\sin^{-1}\frac{x+1}{2}+C,\qquad -3\lt x\lt1$$
Meaning $y=\sqrt{4-t^2}$ is the upper half of a circle of radius $2$. In Chapter 8 the first formula gives the area of a circle, $\pi a^2$, and of an ellipse.
Interactive 2 — Integration by Parts as Two Areas
Plot $v$ against $u$ as $x$ runs from $x_1$ to $x_2$. The red region, between the curve and the $v$-axis, has area $\int u\,dv$; the blue region, between the curve and the $u$-axis, has area $\int v\,du$. Together they fill the big rectangle $u_2v_2$ except for the small one $u_1v_1$. That picture is the formula $\int u\,dv=\big[uv\big]-\int v\,du$. Play stretches the curve; the readouts add the two areas (Section 7.4 makes “area” precise).
red: $\int u\,dv$—
blue: $\int v\,du$—
sum and $u_2v_2-u_1v_1$—
⚠️ Common mistakes when studying alone
1. The wrong first function. If the new integral is harder (a higher power, a log you cannot integrate), you chose the wrong $u$ — go back and swap. ILATE makes the right choice almost every time.
2. A sign error in the second round. $\int u\,dv=uv-\int v\,du$: the minus sign applies to the whole of the next integral, including everything that comes out of it. Use brackets: $x^2e^x-2\left(xe^x-e^x\right)$.
3. Undoing the first step in a cyclic integral. In $\int e^x\sin x\,dx$, if you choose $u=e^x$ the second time after choosing $u=\sin x$ the first time, you get $I=I$. Keep the same kind of first function both times.
4. Forgetting that $\log x$ needs parts. $\int\log x\,dx$ is not $\tfrac1x$ (that is the derivative). It is $x\log x-x+C$.
5. Mixing up the three root formulas. Only $\sqrt{a^2-x^2}$ gives an inverse sine; the other two give logarithms, with a plus sign for $x^2+a^2$ and a minus sign for $x^2-a^2$.
🚀 Beyond the board syllabus (JEE Main · JEE Advanced)
In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. Integration by parts can lower a power step by step, giving a reduction formula. For $I_n=\displaystyle\int\sin^nx\,dx$ with $n\ge2$, write $\sin^nx=\sin^{n-1}x\cdot\sin x$, take $u=\sin^{n-1}x$, and use $\cos^2x=1-\sin^2x$:
$$I_n=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}\,I_{n-2}$$
Worked example. $I_0=x$, so $I_2=-\tfrac12\sin x\cos x+\tfrac12x$, and
$$\int\sin^4x\,dx=-\frac{\sin^3x\cos x}{4}+\frac34I_2=-\frac{\sin^3x\cos x}{4}-\frac{3\sin x\cos x}{8}+\frac{3x}{8}+C$$
The same idea gives $\int x^ne^x\,dx=x^ne^x-n\int x^{n-1}e^x\,dx$, which is Example 2 in compact form.
Exercises
I
Integration by parts. Name the first function by ILATE, then integrate. Check your answers in the lab below.
3 L before A: $u=\log x$, $v=\tfrac{x^3}3$: $\ \tfrac{x^3}{3}\log x-\displaystyle\int\frac{x^2}{3}dx=\frac{x^3}{3}\log x-\frac{x^3}{9}+C$.
4 $u=x$, $v=\tan x$: $\ x\tan x-\displaystyle\int\tan x\,dx$, that is, $x\tan x-\log|\sec x|+C$.
5 Twice: $x^2\sin x-2\displaystyle\int x\sin x\,dx$, and by part 1 this is
$$x^2\sin x+2x\cos x-2\sin x+C$$
6 $u=(\log x)^2$, second function $1$: $x(\log x)^2-2\displaystyle\int\log x\,dx$. With Example 3(a):
$$x(\log x)^2-2x\log x+2x+C$$
7 $u=\tan^{-1}x$, $v=\tfrac{x^2}{2}$: $\tfrac{x^2}{2}\tan^{-1}x-\tfrac12\displaystyle\int\frac{x^2}{1+x^2}dx$, and $\dfrac{x^2}{1+x^2}=1-\dfrac{1}{1+x^2}$. So
$$\frac12\left(x^2+1\right)\tan^{-1}x-\frac x2+C$$
8 $\sin x\cos x=\tfrac12\sin2x$; then $u=x$, $v=-\tfrac12\cos2x$:
$$\frac12\left(-\frac x2\cos2x+\frac14\sin2x\right)+C=-\frac x4\cos2x+\frac18\sin2x+C$$
6 $\dfrac{x-3}{(x-1)^3}=\dfrac{1}{(x-1)^2}-\dfrac{2}{(x-1)^3}$, and $\left(\tfrac{1}{(x-1)^2}\right)'=-\tfrac{2}{(x-1)^3}$: $\ \dfrac{e^x}{(x-1)^2}+C$.
7 As in Example 5: $\ \tfrac12e^x(\sin x+\cos x)+C$.
8 $I=-e^{2x}\cos x+2\displaystyle\int e^{2x}\cos x\,dx$ and $\displaystyle\int e^{2x}\cos x\,dx=e^{2x}\sin x-2I$. So $I=-e^{2x}\cos x+2e^{2x}\sin x-4I$, giving
$$I=\frac{e^{2x}(2\sin x-\cos x)}{5}+C$$
in agreement with the general formula ($a=2$, $b=1$).
III
Square-root integrals. Complete the square where needed and use Theorem 7.10.
4 $u=2x$: $\tfrac12\displaystyle\int\sqrt{1-u^2}\,du=\tfrac12\left(\tfrac u2\sqrt{1-u^2}+\tfrac12\sin^{-1}u\right)$, that is, $\ \dfrac x2\sqrt{1-4x^2}+\dfrac14\sin^{-1}(2x)+C$.
5 $(x+2)^2+2$, so $a^2=2$:
$$\frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left|x+2+\sqrt{x^2+4x+6}\right|+C$$
6 $2-(x-1)^2$, so $a=\sqrt2$:
$$\frac{x-1}{2}\sqrt{1+2x-x^2}+\sin^{-1}\frac{x-1}{\sqrt2}+C$$
7 Split: $\displaystyle\int x\sqrt{x^2+1}\,dx=\tfrac13\left(x^2+1\right)^{3/2}$ (substitute $u=x^2+1$), plus $\displaystyle\int\sqrt{x^2+1}\,dx$ from Theorem 7.10:
$$\tfrac13\left(x^2+1\right)^{3/2}+\tfrac x2\sqrt{x^2+1}+\tfrac12\log\left|x+\sqrt{x^2+1}\right|+C$$
IV
Modelling. Give units and 2 decimal places.
1Drug absorption. A drug enters the blood at the rate $r(t)=50te^{-0.5t}$ mg per hour, starting with none. Find the amount $A(t)$ absorbed after $t$ hours, the amount after $4$ hours, and the total in the long run.
2Driven spring. A mass pushed in time with its own swing has velocity $v(t)=t\cos t$ m/s and starts at $s(0)=0$. Find its displacement $s(t)$ and its value at $t=\pi$ s.
3Curve. A curve through $(1,0)$ has slope $x\log x$ at each point with $x\gt0$. Find its equation and its $y$-value at $x=e$.
💡 Show solutions▶
1 $u=t$, $v=-2e^{-0.5t}$: $\displaystyle\int te^{-0.5t}dt=-2te^{-0.5t}-4e^{-0.5t}$. So $A(t)=-100te^{-0.5t}-200e^{-0.5t}+C$ and $A(0)=0$ gives $C=200$: $A(t)=200-(100t+200)e^{-0.5t}$ mg. $A(4)=200-600e^{-2}\approx118.80$ mg. As $t\to\infty$, $te^{-0.5t}\to0$, so $A\to200$ mg.
2 $s(t)=\displaystyle\int t\cos t\,dt=t\sin t+\cos t+C$ (Example 1), and $s(0)=1+C=0$ gives $C=-1$. $s(\pi)=0-1-1=-2$ m: the mass is $2$ m on the other side of its start.
3 $u=\log x$, $v=\tfrac{x^2}2$: $y=\dfrac{x^2}{2}\log x-\dfrac{x^2}{4}+C$, and $y(1)=-\tfrac14+C=0$ gives $C=\tfrac14$. At $x=e$: $y=\tfrac{e^2}2-\tfrac{e^2}4+\tfrac14=\tfrac{e^2+1}{4}\approx2.10$.
How this is examined: the Board asks for one or two rounds of parts (2–3 marks), an $e^x[f+f']$ integral that needs a little algebra (1–3 marks), and a 5-mark integral of the form $\int(px+q)\sqrt{ax^2+bx+c}\,dx$. JEE disguises the $e^x[f+f']$ pattern and uses cyclic integrals.
E1.Board · 1 mark $\displaystyle\int e^x(\cos x-\sin x)\,dx$ is equal to (A) $e^x\cos x+C$ (B) $e^x\sin x+C$ (C) $-e^x\cos x+C$ (D) $-e^x\sin x+C$
💡 Show solution▶
Idea $f=\cos x$ has $f'=-\sin x$, so the bracket is $f+f'$ (Theorem 7.9).
Answer (A). (B) would need $\sin x+\cos x$; differentiating (C) gives $-e^x(\cos x-\sin x)$, the negative of the integrand.
E2.Board · 2 marks Find $\displaystyle\int x\sec x\tan x\,dx$.
💡 Show solution▶
Parts $u=x$ (A before T), second function $\sec x\tan x$ with integral $\sec x$: $\ x\sec x-\displaystyle\int\sec x\,dx$. (1 mark)
Answer $x\sec x-\log|\sec x+\tan x|+C$. (1 mark)
E3.Board · 3 marks Find $\displaystyle\int x^2\tan^{-1}x\,dx$.
💡 Show solution▶
Parts I before A: $u=\tan^{-1}x$, $v=\tfrac{x^3}{3}$: $\ \dfrac{x^3}{3}\tan^{-1}x-\dfrac13\displaystyle\int\frac{x^3}{1+x^2}\,dx$. (1 mark)
Divide $\dfrac{x^3}{1+x^2}=x-\dfrac{x}{1+x^2}$, whose integral is $\dfrac{x^2}{2}-\dfrac12\log\left(1+x^2\right)$. (1 mark)
E4.Board · 5 marks Find $\displaystyle\int(3x-2)\sqrt{x^2+x+1}\,dx$.
💡 Show solution▶
Split $\dfrac{d}{dx}(x^2+x+1)=2x+1$, and $3x-2=\tfrac32(2x+1)-\tfrac72$. (1 mark)
First part With $u=x^2+x+1$:
$$\tfrac32\int(2x+1)\sqrt{x^2+x+1}\,dx=\tfrac32\cdot\tfrac23u^{3/2}=\left(x^2+x+1\right)^{3/2}$$ (1 mark)
Second part $x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34$, so with $a^2=\tfrac34$ in Theorem 7.10,
$$\int\sqrt{x^2+x+1}\,dx=\frac{2x+1}{4}\sqrt{x^2+x+1}+\frac38\log\left|x+\tfrac12+\sqrt{x^2+x+1}\right|$$ (2 marks)
Answer Multiply the second part by $-\tfrac72$ and add:
$$\left(x^2+x+1\right)^{3/2}-\frac{7(2x+1)}{8}\sqrt{x^2+x+1}-\frac{21}{16}\log\left|x+\tfrac12+\sqrt{x^2+x+1}\right|+C$$ (1 mark)
Algebra $x^2+1=(x+1)^2-2x$, and $\dfrac{x^2+1}{(x+1)^2}=\dfrac{x-1}{x+1}+\dfrac{2}{(x+1)^2}$ (check by adding). With $f=\dfrac{x-1}{x+1}=1-\dfrac{2}{x+1}$, $f'=\dfrac{2}{(x+1)^2}$.
Answer (A), by Theorem 7.9. (C) treats the problem as if only the denominator mattered; (D) integrates as if $e^x$ multiplied a constant.
⏱ Fast route Differentiate the options at $x=0$: the integrand is $1$ there. (A) gives $e^0\left(f(0)+f'(0)\right)=-1+2=1$ ✓; (C) gives $1-2=-1$.
E6.JEE Advanced(Numerical answer.) Let $F$ be the antiderivative of $e^x\sin x$ with $F(0)=0$. Find the value of $2F(\pi)-e^{\pi}$.
💡 Show solution▶
Antiderivative By Example 5, $F(x)=\tfrac12e^x(\sin x-\cos x)+C$, and $F(0)=-\tfrac12+C=0$ gives $C=\tfrac12$.
Answer $2F(\pi)-e^\pi=1$. (A common slip is $C=-\tfrac12$, from losing the sign of $-\cos0$; it gives $-1$.)
🔬 Research–Teaching NexusMini project · about 60 minutes
Research question. After a single tablet, how much of a medicine does the body actually “see”, and how much does the answer change if blood samples are taken every $2$ hours instead of continuously?
Why it matters. Pharmacologists measure a drug's total exposure as the area under its concentration–time curve (“AUC”). Regulators in India, Europe and the USA compare a generic medicine with the original brand by their AUCs. A common model for the blood concentration after an oral dose is $C(t)=Kte^{-kt}$ — exactly a function that integration by parts handles.
What to do.
Take $C(t)=20te^{-0.4t}$ mg/L ($t$ in hours). Find the time and value of the peak concentration by setting $C'(t)=0$.
Use integration by parts to find an antiderivative $F$ of $C$. Compute $F(24)-F(0)$, the exposure over the first day, in mg·h/L.
In a spreadsheet, “sample” $C$ every $2$ hours from $0$ to $24$ and estimate the exposure by adding trapezia $\tfrac12\big(C(t_i)+C(t_{i+1})\big)\cdot2$. Repeat with samples every $1$ hour.
Change $k$ to $0.3$ and $0.5$. How do the peak and the exposure change? Show that the exposure over a very long time is $\dfrac{K}{k^2}$.
Tools: Spreadsheet; your antiderivative from integration by parts.
Hand in: A graph of $C(t)$, your antiderivative, the exact and the two sampled exposures with their percentage errors, and a paragraph on why a clinical trial cannot sample continuously and what that costs in accuracy.
Going further. A slow-release tablet has $C(t)=Kt^2e^{-kt}$. Use parts twice to find its long-run exposure.
Instructor note: $C'(t)=20e^{-0.4t}(1-0.4t)$, so the peak is at $t=2.5$ h with $C=50e^{-1}\approx18.39$ mg/L. $F(t)=-(50t+125)e^{-0.4t}$, so the exposure over $24$ h is $125-1325e^{-9.6}\approx124.91$ mg·h/L. The $2$-hour trapezium sum is about $118.45$ and the $1$-hour sum about $123.26$ (both underestimate, since the curve is mostly concave down near its peak). Long-run exposure $K/k^2=125$; for the slow-release tablet it is $2K/k^3$.
🧪 Self-check quiz
0/5 correct
Q1. To find $\displaystyle\int x\log x\,dx$ by parts, the first function (the one to differentiate) should be