Modules › The definite integral, the Fundamental Theorem and substitution
The definite integral, the Fundamental Theorem and substitution
Riemann sums, definite integral and its properties, FTC parts 1 and 2, substitution rule, average value.
Lecture 7 and the start of Lecture 8 · Tutorial 7 · Definite integrals and their applications
Textbook sections: Thomas ch. 5 Integrals (5.1–5.6) and ch. 6 (6.1–6.4) · Stewart ch. 5 (5.1–5.5), ch. 6 (6.1–6.3, 6.5) and ch. 8 (8.1–8.2)
Textbook sections: Thomas ch. 5 Integrals (5.1–5.6) and ch. 6 (6.1–6.4) · Stewart ch. 5 (5.1–5.5), ch. 6 (6.1–6.3, 6.5) and ch. 8 (8.1–8.2)
The definite integral, the Fundamental Theorem and substitution
Calculus I · Module 8
After this module you can
- Compute left/right/midpoint Riemann sums and recognise a Riemann sum as an integral.
- Apply the properties of the definite integral and the FTC (also with the chain rule).
- Use substitution for indefinite and definite integrals; find areas between curves and average values.
From sums to the integral
- $\int_a^bf\,dx=\lim_{n\to\infty}\sum_{k=1}^nf(c_k)\Delta x$, $\Delta x=\frac{b-a}n$ ($c_k$ = left, right or midpoint).
- $f(x)=x^2-1$ on $[0,2]$, $n=4$: left $-\frac14$, right $\frac74$, midpoint $\frac58$; the exact value is $\frac23$.
- Right sums for $x^2+1$ on $[0,3]$: $\frac{27}{n^3}\frac{n(n+1)(2n+1)}6+3\to12$.
- Recognise $\lim\sum\frac1n\ln\left(1+\frac kn\right)=\int_0^1\ln(1+x)\,dx=2\ln2-1$.
Properties and the FTC
| Statement | Formula |
|---|---|
| reverse limits | $\int_b^af=-\int_a^bf$ |
| additivity | $\int_a^cf=\int_a^bf+\int_b^cf$ |
| average value | $\frac1{b-a}\int_a^bf$ |
| FTC 1 | $\frac{d}{dx}\int_a^xf=f(x)$; with chain rule $\frac{d}{dx}\int_a^{u(x)}f=f(u)u'$ |
| FTC 2 | $\int_a^bf=F(b)-F(a)$ |
$\frac{d}{dx}\int_1^{x^2}\cos t\,dt=2x\cos x^2$ and $\frac{d}{dx}\int_x^5 3t\sin t\,dt=-3x\sin x$.
Substitution
- Indefinite: $u=g(x)$, $du=g'(x)dx$. Definite: change the limits too (or go back to $x$).
- $\int(x^3+x)^5(3x^2+1)dx=\frac{(x^3+x)^6}6+C$; $\int\sqrt{2x+1}\,dx=\frac{(2x+1)^{3/2}}3+C$.
- When $x$ remains: $\int x\sqrt{2x+1}\,dx$ with $u=2x+1$, $x=\frac{u-1}2$: $\frac{(2x+1)^{5/2}}{10}-\frac{(2x+1)^{3/2}}6+C$.
- Always check by differentiating your answer.
Areas between curves
- Area $=\int_a^b(\text{upper}-\text{lower})\,dx$, or integrate in $y$ for $x=\ldots$ curves.
- $x=2y^2$, $x=0$, $y=3$: $\int_0^32y^2dy=18$.
- $y^2=4x$ and $y=4x-2$ meet at $y=-1,2$: area $=\int_{-1}^2\left(\frac{y+2}4-\frac{y^2}4\right)dy=\frac98$.
- Sketch first; split the interval where the curves cross.
Typical mistakes
Limits after substitutionChange the limits or return to $x$ before evaluating.
FTC with a variable limitDo not forget $u'(x)$.
Area vs integralA signed integral can cancel: use $|f|$ or split.
Practice next
- Chapter quizzes Ch. 4 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Riemann sums, trapezoid and Simpson
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Riemann sums (Tutorial 7, Q2)
For $f(x)=x^2+1$ on $[0,3]$, the right-endpoint Riemann sum with $n$ subintervals tends, as $n\to\infty$, to
Show the answer and the solution
Answer: $12$
$\Delta x=\frac3n$, $c_k=\frac{3k}{n}$: $\sum_{k=1}^n\left(\frac{9k^2}{n^2}+1\right)\frac3n=\frac{27}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+3\to9+3=12$ (equal to $\int_0^3(x^2+1)dx=12$).
Worked example 2 · Fundamental Theorem of Calculus (Tutorial 7, Q6)
Find $\dfrac{d}{dt}\displaystyle\int_1^{\sqrt t}\left(x^4+\dfrac{3}{x^3}\right)dx$.
Show the answer and the solution
Answer: $\frac{t^{\frac{7}{2}} + 3}{2 t^{2}}$
FTC-1 with the chain rule: $\frac{d}{dt}\int_1^{u(t)}g=g(u)\,u'$ with $u=\sqrt t$, $u'=\frac1{2\sqrt t}$, $g(u)=u^4+3u^{-3}$: $\left(t^2+3t^{-3/2}\right)\frac{1}{2\sqrt t}=\frac{t^{3/2}}{2}+\frac{3}{2t^2}$.
Worked example 3 · Area between curves (Tutorial 7, Q8(b))
Find the area enclosed by $y^2=4x$ and $y=4x-2$.
Show the answer and the solution
Answer: $\frac{9}{8}$
Intersections: $y^2=4x$ and $y=4x-2\Rightarrow x=\frac{y+2}4$, so $y^2=y+2$, $y=-1,2$. Integrate in $y$: $\int_{-1}^{2}\left(\frac{y+2}{4}-\frac{y^2}{4}\right)dy=\frac98$.
Worked example 4 · Average value (Lecture 7, Mean Value Theorem for integrals)
By the Mean Value Theorem for integrals there is a number $c\in(0,3)$ with $f(c)=\dfrac{1}{b-a}\displaystyle\int_a^bf(x)\,dx$. Find $c$ for $f(x)=x^{2}$.
Show the answer and the solution
Answer: $\sqrt{3}$
$\displaystyle\int_{0}^{3}x^{2}\,dx=9$, so the average value is $\dfrac{1}{3}\cdot9=3$. Solve $f(c)=3$ in $(0,3)$: $c=\sqrt{3}$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Riemann sums, trapezoid and Simpson