Week 9: Integration, Areas & the Fundamental Theorem

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Indefinite integrals, antiderivative rules, Fundamental Theorem of Calculus, signed area vs total area, and u-substitution. # Mathematics I — Week 9: Integration, Areas & the Fundamental Theorem IITM’s Week 9 covers **integration of a function of one variable, computing areas, area under a curve, and the relationshi...

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Mathematics I — Week 9: Integration, Areas & the Fundamental Theorem

IITM’s Week 9 covers integration of a function of one variable, computing areas, area under a curve, and the relationship between derivatives and integrals.

TL;DR

Week 8 asked:
How fast is something changing?
Week 9 asks:
How much change has accumulated?
The central object is the integral.
Two core ideas:
Indefinite integral=antiderivative\boxed{\text{Indefinite integral}=\text{antiderivative}}
and
Definite integral=signed accumulated area\boxed{\text{Definite integral}=\text{signed accumulated area}}
The key bridge is the Fundamental Theorem of Calculus, which links derivatives and integrals.

1. Antiderivative intuition

Suppose:
F′(x)=f(x).F'(x)=f(x).
Then FF is called an antiderivative of ff.
Example:
F(x)=x2.F(x)=x^2.
Then:
F′(x)=2x.F'(x)=2x.
So an antiderivative of:
2x2x
is:
x2.x^2.

2. Indefinite integral

We write:
∫f(x) dx\boxed{ \int f(x)\,dx }
for the family of antiderivatives of ff.
If:
F′(x)=f(x),F'(x)=f(x),
then:
∫f(x) dx=F(x)+C\boxed{ \int f(x)\,dx=F(x)+C }
where:
CC
is an arbitrary constant.

3. Why the constant CC appears

Because derivatives of constants vanish.
For example:
ddx(x2)=2x\frac{d}{dx}(x^2)=2x
but also:
ddx(x2+5)=2x\frac{d}{dx}(x^2+5)=2x
and:
ddx(x2−100)=2x.\frac{d}{dx}(x^2-100)=2x.
So all functions:
x2+Cx^2+C
have derivative:
2x.2x.
Hence:
∫2x dx=x2+C.\int 2x\,dx=x^2+C.

4. Basic integration notation

In:
∫f(x) dx,\int f(x)\,dx,
  • ∫\int = integral sign
  • f(x)f(x) = integrand
  • dxdx = variable of integration
The integral sign is historically a stretched form of the letter SS, referring to a sum.
That becomes intuitive when we discuss area.

5. Power rule for integration

For:
n≠−1,n\neq -1, ∫xn dx=xn+1n+1+C\boxed{ \int x^n\,dx = \frac{x^{n+1}}{n+1}+C }
Compare with differentiation:
ddxxn=nxn−1.\frac{d}{dx}x^n=nx^{n-1}.
Integration roughly reverses the power rule.

6. Examples

∫x2 dx=x33+C\int x^2\,dx = \frac{x^3}{3}+C
because:
ddx(x33)=x2.\frac{d}{dx}\left(\frac{x^3}{3}\right)=x^2.
Similarly:
∫x5 dx=x66+C.\int x^5\,dx = \frac{x^6}{6}+C.

7. Integrating constants

∫c dx=cx+C.\int c\,dx = cx+C.
Why?
Because:
ddx(cx)=c.\frac{d}{dx}(cx)=c.
Example:
∫7 dx=7x+C\boxed{ \int 7\,dx=7x+C }

8. Constant multiple rule

∫cf(x) dx=c∫f(x) dx\boxed{ \int c f(x)\,dx = c\int f(x)\,dx }
Example:
∫5x3 dx=5⋅x44+C\int 5x^3\,dx = 5\cdot\frac{x^4}{4}+C
so:
54x4+C\boxed{ \frac54x^4+C }

9. Sum and difference rules

∫[f(x)+g(x)] dx=∫f(x) dx+∫g(x) dx\boxed{ \int[f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx }
and:
∫[f(x)−g(x)] dx=∫f(x) dx−∫g(x) dx\boxed{ \int[f(x)-g(x)]\,dx = \int f(x)\,dx-\int g(x)\,dx }

10. Polynomial integration

Example:
∫(3x2−4x+5) dx.\int(3x^2-4x+5)\,dx.
Integrate term by term:
=3x33−4x22+5x+C.=3\frac{x^3}{3} -4\frac{x^2}{2} +5x+C.
Simplify:
x3−2x2+5x+C\boxed{ x^3-2x^2+5x+C }

11. Check your answer by differentiating

If you think:
∫f(x) dx=F(x)+C,\int f(x)\,dx=F(x)+C,
differentiate:
F′(x).F'(x).
If:
F′(x)=f(x),F'(x)=f(x),
your integration is correct.
This is one of the best self-checks in calculus.

12. Special case: ∫1x dx\int \frac1x\,dx

The ordinary power rule fails for:
n=−1n=-1
because it would require division by:
n+1=0.n+1=0.
Instead:
∫1x dx=ln⁡∣x∣+C\boxed{ \int\frac1x\,dx = \ln|x|+C }
for:
x≠0.x\neq0.
Why absolute value?
Because:
ddxln⁡∣x∣=1x\frac{d}{dx}\ln|x|=\frac1x
on both positive and negative domains.

13. Exponential integrals

Because:
ddxex=ex,\frac{d}{dx}e^x=e^x,
we have:
∫ex dx=ex+C\boxed{ \int e^x\,dx=e^x+C }
For general base:
∫ax dx=axln⁡a+C\boxed{ \int a^x\,dx = \frac{a^x}{\ln a}+C }
for:
a>0,a≠1.a>0,\quad a\neq1.

14. Integration as accumulated area

Now the second interpretation.
Suppose:
f(x)≥0f(x)\geq0
on:
[a,b].[a,b].
The area under:
y=f(x)y=f(x)
from x=ax=a to x=bx=b is represented by:
∫abf(x) dx\boxed{ \int_a^b f(x)\,dx }
This is a definite integral.

15. Definite integral notation

∫abf(x) dx\int_a^b f(x)\,dx
where:
  • aa = lower limit
  • bb = upper limit
  • f(x)f(x) = integrand
Unlike the indefinite integral, the definite integral evaluates to a number.

Imagine dividing:
[a,b][a,b]
into many thin rectangles.
For each rectangle:
area≈height×width.\text{area}\approx \text{height}\times\text{width}.
Then add all rectangle areas:
∑f(xi)Δx.\sum f(x_i)\Delta x.
As the rectangles become thinner:
Δx→0,\Delta x\to0,
the approximation becomes exact.
This limiting sum gives:
∫abf(x) dx\boxed{ \int_a^b f(x)\,dx }
This is the conceptual origin of integration.

17. Riemann sum intuition

A typical approximation is:
∑i=1nf(xi∗)Δx.\sum_{i=1}^n f(x_i^*)\Delta x.
As:
n→∞,n\to\infty,
the width:
Δx→0.\Delta x\to0.
Then:
∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗)Δx\boxed{ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\Delta x }
under appropriate conditions.
This ties Week 7 limits directly into Week 9 integration.

18. Fundamental Theorem of Calculus

This is one of the most important results in calculus.
If:
F′(x)=f(x),F'(x)=f(x),
then:
∫abf(x) dx=F(b)−F(a)\boxed{ \int_a^b f(x)\,dx = F(b)-F(a) }
This means we do not usually need to compute infinite rectangle sums manually.
Instead:
  1. find an antiderivative
  2. evaluate at upper bound
  3. subtract value at lower bound

19. Example

Compute:
∫02x dx.\int_0^2 x\,dx.
An antiderivative is:
F(x)=x22.F(x)=\frac{x^2}{2}.
So:
∫02x dx=[x22]02.\int_0^2x\,dx = \left[\frac{x^2}{2}\right]_0^2.
Evaluate:
=222−022= \frac{2^2}{2} - \frac{0^2}{2} =2.=2.
Therefore:
2\boxed{2}

20. Bracket notation

We often write:
[F(x)]ab\left[F(x)\right]_a^b
to mean:
F(b)−F(a).F(b)-F(a).
So:
[x33]14\left[\frac{x^3}{3}\right]_1^4
means:
433−133.\frac{4^3}{3}-\frac{1^3}{3}.

21. No +C+C in a definite integral

When evaluating:
∫abf(x) dx,\int_a^b f(x)\,dx,
we do not need to write:
+C.+C.
Why?
Suppose antiderivative is:
F(x)+C.F(x)+C.
Then:
[F(b)+C]−[F(a)+C][F(b)+C]-[F(a)+C]
equals:
F(b)−F(a).F(b)-F(a).
The constants cancel.

22. Example: polynomial definite integral

Evaluate:
∫13(2x+1) dx.\int_1^3(2x+1)\,dx.
Antiderivative:
x2+x.x^2+x.
Thus:
[x2+x]13.[x^2+x]_1^3.
At 33:
9+3=12.9+3=12.
At 11:
1+1=2.1+1=2.
Therefore:
10\boxed{10}

23. Geometric verification

For:
f(x)=2x+1f(x)=2x+1
between:
x=1x=1
and:
x=3,x=3,
the graph forms a trapezoid.
Heights:
f(1)=3f(1)=3
and:
f(3)=7.f(3)=7.
Width:
2.2.
Trapezoid area:
12(3+7)(2)=10.\frac12(3+7)(2)=10.
Matches the integral.

24. Signed area

This is crucial:
A definite integral gives signed area, not always ordinary geometric area.
If:
f(x)>0,f(x)>0,
contribution is positive.
If:
f(x)<0,f(x)<0,
contribution is negative.
So regions below the xx-axis subtract.

25. Example below the axis

Consider:
f(x)=−2f(x)=-2
on:
[0,3].[0,3].
Then:
∫03−2 dx=−6.\int_0^3 -2\,dx = -6.
Geometric area is:
6,6,
but definite integral is:
−6.-6.

26. Net area vs total area

Suppose a graph crosses the xx-axis.
Then:
∫abf(x) dx\int_a^b f(x)\,dx
gives net signed area.
If the question asks for actual geometric area, you need:
∫ab∣f(x)∣ dx\boxed{ \int_a^b |f(x)|\,dx }
or split at the zeros.

27. Example: signed cancellation

Take:
f(x)=xf(x)=x
on:
[−1,1].[-1,1].
Then:
∫−11x dx=0.\int_{-1}^{1}x\,dx=0.
Why?
Area below axis from −1-1 to 00 is:
−12.-\frac12.
Area above axis from 00 to 11 is:
+12.+\frac12.
They cancel.
But total geometric area is:
1.1.

28. Symmetry shortcut: odd functions

If:
f(−x)=−f(x),f(-x)=-f(x),
then ff is odd.
Over symmetric limits:
[−a,a],[-a,a],
we get:
∫−aaf(x) dx=0\boxed{ \int_{-a}^a f(x)\,dx=0 }
Example:
f(x)=x3.f(x)=x^3.
Thus:
∫−22x3 dx=0.\int_{-2}^2x^3\,dx=0.

29. Symmetry shortcut: even functions

If:
f(−x)=f(x),f(-x)=f(x),
then ff is even.
Then:
∫−aaf(x) dx=2∫0af(x) dx\boxed{ \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx }
Example:
f(x)=x2.f(x)=x^2.

30. Basic properties of definite integrals

Same limits

∫aaf(x) dx=0\boxed{ \int_a^a f(x)\,dx=0 }

Reversing limits

∫abf(x) dx=−∫baf(x) dx\boxed{ \int_a^b f(x)\,dx = -\int_b^a f(x)\,dx }

Splitting interval

For any cc between aa and bb:
∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\boxed{ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx }

31. Linearity of definite integrals

∫ab[αf(x)+βg(x)] dx=α∫abf(x) dx+β∫abg(x) dx\boxed{ \int_a^b [\alpha f(x)+\beta g(x)]\,dx = \alpha\int_a^b f(x)\,dx + \beta\int_a^b g(x)\,dx }
This is extremely useful.

32. Area between two curves

Suppose:
f(x)≥g(x)f(x)\geq g(x)
on:
[a,b].[a,b].
Then the area between the curves is:
∫ab[f(x)−g(x)] dx\boxed{ \int_a^b [f(x)-g(x)]\,dx }
Think:
top−bottom.\text{top} - \text{bottom}.

33. Example: area between curves

Find the area between:
y=xy=x
and:
y=x2y=x^2
from:
x=0x=0
to:
x=1.x=1.
On this interval:
x≥x2.x\geq x^2.
So area:
∫01(x−x2) dx.\int_0^1(x-x^2)\,dx.
Integrate:
[x22−x33]01.\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1.
Thus:
12−13=16.\frac12-\frac13 = \boxed{\frac16}.

34. Find intersection points first

If bounds are not given, solve:
f(x)=g(x)f(x)=g(x)
to find where curves intersect.
Example:
x=x2.x=x^2.
Then:
x2−x=0x^2-x=0 x(x−1)=0.x(x-1)=0.
So:
x=0,1.x=0,\quad1.
These become integration limits.

35. Antiderivative families

Suppose:
f′(x)=6x.f'(x)=6x.
Then:
f(x)=3x2+C.f(x)=3x^2+C.
If we additionally know:
f(1)=10,f(1)=10,
then:
3(1)2+C=10.3(1)^2+C=10.
So:
C=7.C=7.
Thus:
f(x)=3x2+7\boxed{ f(x)=3x^2+7 }
An initial condition picks one member from the antiderivative family.

36. Recovering position from velocity

Suppose velocity:
v(t)=s′(t).v(t)=s'(t).
Then displacement from:
t=at=a
to:
t=bt=b
is:
∫abv(t) dt=s(b)−s(a)\boxed{ \int_a^b v(t)\,dt = s(b)-s(a) }
This gives the physical meaning of integration as accumulation.

37. Displacement vs distance

If velocity becomes negative, then:
∫abv(t) dt\int_a^b v(t)\,dt
gives net displacement.
Total distance traveled requires accounting for magnitude:
∫ab∣v(t)∣ dt.\int_a^b |v(t)|\,dt.
Exactly analogous to signed area vs geometric area.

38. Fundamental Theorem — deeper intuition

Suppose:
A(x)=∫axf(t) dt.A(x)=\int_a^x f(t)\,dt.
This defines accumulated area from aa to xx.
Then, under suitable conditions:
A′(x)=f(x)\boxed{ A'(x)=f(x) }
So:
the instantaneous rate at which accumulated area changes is the current height of the function.
That is the deeper reason differentiation and integration are inverses.

39. Why does A′(x)=f(x)A'(x)=f(x) make sense?

Increase xx by a tiny amount:
Δx.\Delta x.
The new area added is approximately a thin rectangle:
ΔA≈f(x)Δx.\Delta A\approx f(x)\Delta x.
Thus:
ΔAΔx≈f(x).\frac{\Delta A}{\Delta x}\approx f(x).
As:
Δx→0,\Delta x\to0,
we get:
A′(x)=f(x).A'(x)=f(x).
This is one of the nicest intuitions in calculus.

40. Substitution intuition

Sometimes an integral contains a nested expression.
Example:
∫2x(x2+1)3 dx.\int 2x(x^2+1)^3\,dx.
Notice:
ddx(x2+1)=2x.\frac{d}{dx}(x^2+1)=2x.
Let:
u=x2+1.u=x^2+1.
Then:
du=2x dx.du=2x\,dx.
So the integral becomes:
∫u3 du.\int u^3\,du.
Thus:
u44+C.\frac{u^4}{4}+C.
Substitute back:
(x2+1)44+C\boxed{ \frac{(x^2+1)^4}{4}+C }
This is uu-substitution, essentially the reverse chain rule.

41. Reverse chain-rule mental model

Differentiation:
ddxF(g(x))=F′(g(x))g′(x).\frac{d}{dx}F(g(x)) = F'(g(x))g'(x).
Integration reverses that pattern:
∫F′(g(x))g′(x) dx=F(g(x))+C\boxed{ \int F'(g(x))g'(x)\,dx = F(g(x))+C }
This is the cleanest way to understand substitution.

42. Example

Evaluate:
∫3x2ex3 dx.\int 3x^2e^{x^3}\,dx.
Let:
u=x3.u=x^3.
Then:
du=3x2 dx.du=3x^2\,dx.
So:
∫eu du=eu+C.\int e^u\,du = e^u+C.
Therefore:
ex3+C\boxed{ e^{x^3}+C }

43. Another example

∫2xx2+1 dx.\int\frac{2x}{x^2+1}\,dx.
Let:
u=x2+1.u=x^2+1.
Then:
du=2x dx.du=2x\,dx.
So:
∫1u du=ln⁡∣u∣+C.\int\frac1u\,du = \ln|u|+C.
Since:
x2+1>0,x^2+1>0,
we get:
ln⁡(x2+1)+C\boxed{ \ln(x^2+1)+C }

44. Definite integral with substitution

Consider:
∫012x(x2+1) dx.\int_0^1 2x(x^2+1)\,dx.
Let:
u=x2+1.u=x^2+1.
When:
x=0,x=0, u=1.u=1.
When:
x=1,x=1, u=2.u=2.
So:
∫12u du.\int_1^2u\,du.
Then:
[u22]12=2−12=32.\left[\frac{u^2}{2}\right]_1^2 = 2-\frac12 = \boxed{\frac32}.

45. Average value of a function

The average value of:
f(x)f(x)
on:
[a,b][a,b]
is:
favg=1b−a∫abf(x) dx\boxed{ f_{\text{avg}} = \frac1{b-a} \int_a^b f(x)\,dx }
This generalizes the ordinary idea:
average=totalamount.\text{average} = \frac{\text{total}}{\text{amount}}.

46. Example: average value

For:
f(x)=xf(x)=x
on:
[0,2],[0,2], favg=12∫02x dx.f_{\text{avg}} = \frac12 \int_0^2x\,dx.
We know:
∫02x dx=2.\int_0^2x\,dx=2.
Therefore:
favg=1\boxed{ f_{\text{avg}}=1 }
which matches the intuitive average height.

Common mistake: forgetting
+C+C
For:
∫f(x) dx,\int f(x)\,dx,
you generally need:
+C.+C.
For definite integrals, you do not.

Common mistake: integrating powers incorrectly
Wrong:
∫x3 dx=3x2.\int x^3\,dx=3x^2.
That is differentiation-style thinking.
Correct:
∫x3 dx=x44+C\boxed{ \int x^3\,dx=\frac{x^4}{4}+C }
Integration:
add 11 to exponent, then divide by new exponent.

Common mistake: power rule for
x−1x^{-1}
Do not use:
xn+1n+1\frac{x^{n+1}}{n+1}
when:
n=−1.n=-1.
Instead:
∫x−1 dx=ln⁡∣x∣+C\boxed{ \int x^{-1}\,dx = \ln|x|+C }
Common mistake: integral equals geometric area
Not always.
∫abf(x) dx\int_a^b f(x)\,dx
is signed.
If ff crosses below the axis, negative parts subtract.
If question says total area, split intervals or use absolute values.

Common mistake: upper minus lower backwards
Correct:
F(b)−F(a)\boxed{ F(b)-F(a) }
not:
F(a)−F(b).F(a)-F(b).
Common mistake: integrating products termwise
There is no general rule:
∫f(x)g(x) dx=(∫f)(∫g).\int f(x)g(x)\,dx = \left(\int f\right) \left(\int g\right).
That is false.
Products may require:
  • substitution
  • other methods later
So recognize structure instead of inventing a product-integration rule.

Common mistake: confusing derivative and integral symbols
Derivative:
ddxf(x)\frac{d}{dx}f(x)
asks for rate of change.
Integral:
∫f(x) dx\int f(x)\,dx
asks for antiderivative/accumulation.
They are linked but not the same operation.

54. Derivative ↔ integral table

DifferentiationIntegration
local changeaccumulated change
tangent slopesigned area
reduces powersincreases powers
F→F′F\to F'f→Ff\to F
derivative of constant = 00integral introduces +C+C
chain rulesubstitution reverses chain rule
This table captures most of the conceptual bridge.

55. Integration formula sheet

Power rule

For n≠−1n\neq-1:
∫xn dx=xn+1n+1+C\boxed{ \int x^n\,dx = \frac{x^{n+1}}{n+1}+C }

Constant

∫c dx=cx+C\boxed{ \int c\,dx=cx+C }

Reciprocal

∫1x dx=ln⁡∣x∣+C\boxed{ \int\frac1x\,dx = \ln|x|+C }

Exponential

∫ex dx=ex+C\boxed{ \int e^x\,dx=e^x+C }

General exponential

∫ax dx=axln⁡a+C\boxed{ \int a^x\,dx = \frac{a^x}{\ln a}+C }

56. Definite integral formula sheet

Fundamental theorem

If:
F′=f,F'=f,
then:
∫abf(x) dx=F(b)−F(a)\boxed{ \int_a^b f(x)\,dx = F(b)-F(a) }

Reverse bounds

∫abf=−∫baf\boxed{ \int_a^b f = -\int_b^a f }

Split intervals

∫abf=∫acf+∫cbf\boxed{ \int_a^b f = \int_a^c f+\int_c^b f }

Area between curves

A=∫ab(top−bottom) dx\boxed{ A= \int_a^b (\text{top}-\text{bottom})\,dx }

57. Full worked example

Evaluate:
∫02(x2+2x+1) dx.\int_0^2(x^2+2x+1)\,dx.
Antiderivative:
F(x)=x33+x2+x.F(x) = \frac{x^3}{3}+x^2+x.
Therefore:
[x33+x2+x]02.\left[ \frac{x^3}{3}+x^2+x \right]_0^2.
At:
x=2:x=2: 83+4+2=83+6=263.\frac83+4+2 = \frac83+6 = \frac{26}{3}.
At:
x=0:x=0: 0.0.
So:
263\boxed{ \frac{26}{3} }

58. Full area example crossing the axis

Find total area between:
y=xy=x
and the xx-axis from:
x=−2x=-2
to:
x=3.x=3.
The function changes sign at:
x=0.x=0.
So total area:
−∫−20x dx+∫03x dx.-\int_{-2}^{0}x\,dx + \int_0^3x\,dx.
First:
∫−20x dx=[x22]−20=0−2=−2.\int_{-2}^{0}x\,dx = \left[\frac{x^2}{2}\right]_{-2}^0 = 0-2 = -2.
Magnitude:
2.2.
Second:
∫03x dx=92.\int_0^3x\,dx = \frac92.
Therefore total geometric area:
2+92=132.2+\frac92 = \boxed{\frac{13}{2}}.

59. Weeks 7–9 as one calculus story

text
Week 7
LIMITS
"What happens as we approach?"
        ↓
Week 8
DERIVATIVE
"How fast is it changing now?"
        ↓
Week 9
INTEGRAL
"How much has accumulated?"
And the Fundamental Theorem says:
differentiation and integration are inverse processes\boxed{ \text{differentiation and integration are inverse processes} }
That is the central conceptual payoff of the calculus portion so far.

60. What you should actually master

By the end of Week 9, you should be able to:
  1. Explain what an antiderivative is.
  2. Understand why indefinite integrals include +C+C.
  3. Apply the integration power rule.
  4. Integrate polynomials.
  5. Handle: ∫1x dx.\int \frac1x\,dx.
  6. Integrate basic exponentials.
  7. Interpret definite integrals as signed accumulation.
  8. Use: F(b)−F(a).F(b)-F(a).
  9. Distinguish signed area from geometric area.
  10. Split integrals at zeros/intersections.
  11. Compute area between two simple curves.
  12. Understand the Fundamental Theorem of Calculus.
  13. Recognize basic substitution as the reverse chain rule.
  14. Recover a function from its derivative plus an initial condition.
  15. Interpret integrals as displacement/accumulation.
If those are comfortable, Week 9 is under control.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Fundamental Theorem of Calculus (Part 1) :: ddx[∫axf(t)dt]=f(x)\frac{d}{dx} \left[\int_a^x f(t) dt\right] = f(x)
  • Fundamental Theorem of Calculus (Part 2) :: ∫abf(x)dx=F(b)−F(a)\int_a^b f(x) dx = F(b) - F(a), where F′(x)=f(x)F'(x) = f(x)
  • Power rule for integration ∫xndx\int x^n dx (n≠−1n \neq -1) :: xn+1n+1+C\frac{x^{n+1}}{n+1} + C
  • Indefinite integral ∫1xdx\int \frac{1}{x} dx :: ln⁡∣x∣+C\ln|x| + C
  • Indefinite integral ∫ekxdx\int e^{kx} dx :: 1kekx+C\frac{1}{k} e^{kx} + C
  • Geometric meaning of ∫abf(x)dx\int_a^b f(x) dx :: Net signed area between curve y=f(x)y = f(x) and x-axis
  • u-substitution is the integral counterpart of :: The Chain Rule in differentiation

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

Evaluate the definite integral: ∫02(3x2−4x+1)dx\int_0^2 (3x^2 - 4x + 1) dx.
NAT
3 Unit Assessment
2

Question 2

What is the derivative with respect to xx of g(x)=∫1x1+t4dtg(x) = \int_1^x \sqrt{1 + t^4} dt?
MCQ
3 Unit Assessment
A. 1+x4\sqrt{1 + x^4}
B. 4x321+x4\frac{4x^3}{2\sqrt{1 + x^4}}
C. 1+x4−2\sqrt{1 + x^4} - \sqrt{2}
D. 1+x41 + x^4

Active Recall Session

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