# How to Integrate y with respect to x in Calculus

Knowing how to integrate y with respect to x using calculus will help you to solve math problems and physics problems. Geometrical problems and problems about moving objects can be solved by integrating y with respect to x.

## What is Integration?

Integration is:

- Finding the area under the curve. When you integrate y with respect to x, you are finding the area under the curve of y plotted against x.
- The opposite of differentiation. If you know how to differentiate, then you already know how to integrate!

## How to Integrate y with respect to x

To integrate y with respect to x, you need to find a function that differentiates with respect to x to give y.

The rules of integration are opposite to the rules of differentiation.

For example...

**How to differentiate** a power function, x^{n}

- Multiply by the original exponent (n)
- Reduce the exponent by one (to give n-1):

d/dx (x^{n}) = n x^{n-1}

**How to integrate** a power function, x^{n}

(You just do the opposite steps, in the opposite order...)

- Increase the exponent (n) by one (to give n+1)
- Divide by the new exponent (n+1)

∫ x^{n} = x^{n+1} / (n+1) + c

Note: we added a constant "+c" to the result. This is because constants disappear under differentiation.

## Quick Reference: Integration

y = x^{n} --------------> x^{n+1} / (n+1)

y = constant ------> x

y = e^{x} ---------------> e^{x}

y = 1/x ------------> ln(x)

y = sin(x) -----------> -cos(x)

y = cos(x) ----------> sin(x)

## Integration with Limits

Sometimes, you will be asked to integrate y with respect to x between two limits. This means finding the area under the curve y = f(x) between the two given values of x.

**How to Integrate y with Respect to x Between Limits**

- Find the integral of y with respect to x as shown above. For more complicated functions, use the Quick Reference box.
- Instead of adding "+c" to your answer, substitute the higher value of x into your result. Do this again for the lower value. Subtract one from the other to find the difference between them.

*Example: Integrate y = x ^{2} + 1 with respect to x, for x between 0 and 1.*

- Increase the exponent of each term by one, and divide each term by the new exponent ∫ y dx = ∫ (x
^{2}+ 1) dx = x^{3}/ 3 + x - At x = 1, x
^{3}/ 3 + x = 4/3. At x = 0, x^{3}/ 3 + x = 0 - Find the difference between your results = 4/3 - 0 = 4/3.

*Solution: The integral of y=x ^{2} + 1 for x between 0 and 1 is 4/3.*