Calculus is the study of the changes taking place in a system over time. Mathematicians, scientists, and engineers use calculus in their work to make predictions based on the information they have. Calculus allows us to calculate change, even when it's not constant, which is usually the case in real-life situations. It is a very practical branch of mathematics and finds application in fields as diverse as science and technology, economics and statistics, and medicine and demography. There are two branches of calculus- differential calculus and integral calculus.

## Basics of Calculus

### Solved Example

**Question:**Solve $\int_{1}^{2}x^5dx$

**Solution:**

$\int_{1}^{2}x^5dx$ = [$\frac{x^6}{6}]_{1}^{2}$

[ $\int x^n$ = $\frac{x^{n +1}}{n + 1}$]

= $\frac{1}{6}$ [$2^6 - 1^6$]

= $\frac{1}{6}$ [64 - 1]

= $\frac{1}{6}$ * 63

= 10.5

[ $\int x^n$ = $\frac{x^{n +1}}{n + 1}$]

= $\frac{1}{6}$ [$2^6 - 1^6$]

= $\frac{1}{6}$ [64 - 1]

= $\frac{1}{6}$ * 63

= 10.5

## Calculus Help

### Solved Examples

**Question 1:**Evaluate $\int_{0}^{1}$ $\frac{(tan^{-1}z)^2}{1 + z^2}$ dz

**Solution:**

Given: $\int_{0}^{1}$ $\frac{(tan^{-1}z)^2}{1 + z^2}$ dz

Substitution Method:

Put $tan^{-1}z$ = t

By differentiating, we get

$\frac{1}{1 + z^2}$ dz = dt

When z = 0, t = $tan^{-1}$0 = 0

When z = 1, t = $tan^{-1}$1 = $\frac{\pi}{4}$

Now

$\int_{0}^{\frac{\pi}{4}}$ $t^2 dt$ = [$\frac{t^3}{3}]_{0}^{\frac{\pi}{4}}$

= $\frac{1}{3}$[($\frac{\pi}{4})^3$]

= $\frac{\pi^3}{192}$

Substitution Method:

Put $tan^{-1}z$ = t

By differentiating, we get

$\frac{1}{1 + z^2}$ dz = dt

When z = 0, t = $tan^{-1}$0 = 0

When z = 1, t = $tan^{-1}$1 = $\frac{\pi}{4}$

Now

$\int_{0}^{\frac{\pi}{4}}$ $t^2 dt$ = [$\frac{t^3}{3}]_{0}^{\frac{\pi}{4}}$

= $\frac{1}{3}$[($\frac{\pi}{4})^3$]

= $\frac{\pi^3}{192}$

**Question 2:**Solve the differential equation $\frac{dy}{dx}$ + $\frac{y}{x}$ = sin x

**Solution:**

Given differential equation is $\frac{dy}{dx}$ + $\frac{y}{x}$ = sin x

Integrating factor (IF) = $e^{\int Pdx}$

Then the solution is y * IF = $\int Q * IF$ $dx$

By comparing given and general equation, we have

P = $\frac{1}{x}$ and Q = sin x

IF = $e^{\int (\frac{1}{x})dx}$ = $e^{log\ x}$ = x

The solution for the given differential equation is

y * x = $\int sin x * x$ $dx$

= -x cos x + sin x + c (Integrating by parts)

=> y = - cos x + $\frac{sin\ x}{x}$ + $\frac{c}{x}$.

**The general differential equation $\frac{dy}{dx}$ + Py = Q**

Step 1:Step 1:

Integrating factor (IF) = $e^{\int Pdx}$

Then the solution is y * IF = $\int Q * IF$ $dx$

By comparing given and general equation, we have

P = $\frac{1}{x}$ and Q = sin x

**Step 2**: Find IFIF = $e^{\int (\frac{1}{x})dx}$ = $e^{log\ x}$ = x

**:**

Step 3Step 3

The solution for the given differential equation is

y * x = $\int sin x * x$ $dx$

= -x cos x + sin x + c (Integrating by parts)

=> y = - cos x + $\frac{sin\ x}{x}$ + $\frac{c}{x}$.