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Understanding Calculus

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Integral Calculus

What Is Integration?

Integration is just:

👉 Adding tiny pieces together to get the whole.

That’s it.
Not demons.
Not wizard math.
Just adding small bits.

Imagine This Example

You have a bucket.
Someone pours water into it very slowly, drop by drop.

You want to know:

How much water is inside after some time?

You can’t count each drop one by one — too many.

So you add all the tiny drops together.

That “adding small, small pieces” is integration.

See It Another Way

Imagine you’re painting a wall with tiny dots

Each dot is small, but together they fill the whole wall.

Integration = adding all the dots.

How Is Integration Connected to Differentiation?

Differentiation breaks things into tiny pieces (rates).
Integration puts tiny pieces back together.

Differentiation = breaking

Integration = rebuilding

They’re opposites — like tying and untying shoelaces.

The Symbol for Integration

∫\int

That long S-shaped symbol ∫ means:

“Sum up everything.”

Differentiation helps us describe how fast something is changing (rate of change). Anti-differentiation (integration) helps us find the original quantity or the total accumulated amount from that rate.

So, if differentiation answers “How fast is it changing?”,

Anti-differentiation (Integration) answers

“How much has changed in total?” and “What original function produced this rate?”

Meaning of Anti-differentiation

If a function F(x) is such that 

ddx[F(x)]=f(x) \frac{d}{dx}[F(x)] = f(x)

then F(x) is an antiderivative of f(x).

We write the indefinite integral as:

∫f(x)dx=F(x)+C∫ f(x) dx = F(x) + C 

where C is the constant of integration.

Why do we add C?

Because many different functions have the same derivative.

For instance,

ddx(x2)=2x \frac{d}{dx}(x²) = 2x

and

ddx(x2+7)=2x \frac{d}{dx} (x²+7) = 2x

So the antiderivative of 2x is .

x2+Cx² + C

Basic Rules (Indefinite Integrals)

1. Power Rule

∫xndx=+C(n≠−1)∫ xⁿ dx =   +  C   (n ≠ −1)

 Add 1 to the power

Divide by the new power

Add + C

xn+1n+1dx+C\frac{x^{n+1}}{n + 1}dx + C

2. Constant Multiple Rule

∫kf(x)dx=k∫f(x)dx∫ k f(x) dx = k ∫ f(x) dx

3. Sum/Difference Rule

∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

4. Important Special Case

∫1xdx=ln|x|+C∫ \frac{1}{x} dx = ln|x| + C

Definite Integrals and Total Accumulation

∫abf(x)dx=F(b)−F(a)∫ₐᵇ f(x) dx = F(b) − F(a)

This is the Fundamental Theorem of Calculus. It gives the total accumulated amount of f(x) from a to b. In many cases, it also represents area under the curve, but the “area” can mean real scientific quantities.

Uses of Integration

Integration helps to:

find area under curves

find distance traveled

calculate volume

solve physics problems

Example 1

Evaluate:

∫(6x2−4x+9)dx ∫ (6x^2 − 4x + 9) dx

Solution

∫(6x2−4x+9)dx∫ (6x^2 − 4x + 9) dx
=∫6x2dx−∫4xdx+∫9dx= ∫ 6x^2 dx − ∫ 4x dx + ∫ 9 dx
=6.x33−4.x22+9x\frac{}{} = 6.\frac{x^3}{3} − 4.\frac{x^2}{2} + 9x
=2x3−2x2+9x+C= 2x^3 – 2x^2 + 9x + C

Example 2

Evaluate:

∫(5x4+3x−7)dx ∫ (5x^4 + 3x − 7) dx

Solution

∫(5x4+3x−7)dx∫ (5x^4 + 3x − 7) dx
∫5x4dx+∫3xdx−∫7dx∫ 5x^4 dx + ∫ 3x dx − ∫ 7 dx
5.x55+3x22–7+C5.\frac{x^5}{5} + 3\frac{x^2}{2} – 7 + C
x5+3x22−7x+Cx^5 + \frac{3x^2}{2} – 7x + C

Example 3

Evaluate:∫(3x+2x3)dxEvaluate: ∫ (\frac{3}{x} + 2x^3) dx

Solution:

∫(3x+2x3)dx∫ (\frac{3}{x} + 2x^3) dx
3∫1xdx+3∫x3dx3∫ \frac{1}{x} dx + 3∫ x^3dx
3ln|x|+x44+C3 ln|x| + \frac{x^4}{4} + C
3ln|x|+x44+C3 ln|x| + \frac{x^4}{4} + C

Example 4

Evaluate

∫251xdx\int_{2}^{5} \frac{1}{x}\, dx

Solution

∫1xdx=ln⁡x+C\int \frac{1}{x}\, dx = \ln x + C
∫251xdx\int_{2}^{5} \frac{1}{x}\, dx
=[ln⁡x]25= \left[ \ln x \right]_{2}^{5}
=ln⁡5−ln⁡2= \ln 5 – \ln 2
=ln⁡(52)= \ln\left( \frac{5}{2} \right)

Example 5

∫0πsin⁡xdx\int_{0}^{\pi} \sin x \, dx

Solution

∫sin⁡xdx=−cos⁡x+C\int \sin x\, dx = -\cos x + C
∫0πsin⁡xdx\int_{0}^{\pi} \sin x\, dx
=[−cos⁡x]0π= \left[ -\cos x \right]_{0}^{\pi}
=(−cos⁡π)−(−cos⁡0)= \left( -\cos \pi \right) – \left( -\cos 0 \right)
=−(−1)−(−1)= -(-1) – ( -1 )
22

Example 6

∫14xdx\int_{1}^{4} x\, dx

Solution

∫xdx\int x\, dx
=x22+C= \frac{x^2}{2} + C
∫14xdx\int_{1}^{4} x\, dx
=[x22]14= \left[ \frac{x^2}{2} \right]_{1}^{4}
=162−12= \frac{16}{2} – \frac{1}{2}
=8−12= 8 – \frac{1}{2}
=152= \frac{15}{2}

Example 7

Find the area under the curve

y=4−x2fromx=0tox=2y = 4 – x^2 from x = 0 to x = 2

Solution

Area=∫02(4−x2)dx\text{Area} = \int_{0}^{2} (4 – x^2)\, dx
∫(4−x2)dx\int (4 – x^2)\, dx
=4x−x33= 4x – \frac{x^3}{3}
=[4x−x33]02= \left[4x – \frac{x^3}{3} \right]_{0}^{2}
=(8−83)−(0−0)= \left( 8 – \frac{8}{3} \right) – (0 – 0)
=243−83= \frac{24}{3} – \frac{8}{3}
=163= \frac{16}{3}

Example 8

A particle moves with velocity

v(t)=3t2+2t.v(t)=3t ^2 +2t.
Find the distance travelledFind\ the\ distance\ travelled
 from t=0 to t=3.\ from\ t=0\ to\ t=3.

Solution

Distance=∫03(3t2+2t)dt\text{Distance} = \int_{0}^{3} (3t^2 + 2t)\, dt
∫(3t2+2t)dt\int (3t^2 + 2t)\, dt
=t3+t2= t^3 + t^2
Distance=[t3+t2]03\text{Distance} = \left[ t^3 + t^2 \right]_{0}^{3}
=(27+9)−(0+0)= (27 + 9) – (0 + 0)
=36 units= 36 \text{ units}

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