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, xn

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

d/dx (xn) = n xn-1

How to integrate a power function, xn

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

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

∫ xn = xn+1 / (n+1) + c

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

Quick Reference: Integration

y = xn --------------> xn+1 / (n+1)

y = constant ------> x

y = ex ---------------> ex

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

1. Find the integral of y with respect to x as shown above. For more complicated functions, use the Quick Reference box.
2. 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 = x2 + 1 with respect to x, for x between 0 and 1.

1. Increase the exponent of each term by one, and divide each term by the new exponent ∫ y dx = ∫ (x2 + 1) dx = x3 / 3 + x
2. At x = 1, x3 / 3 + x = 4/3. At x = 0, x3 / 3 + x = 0
3. Find the difference between your results = 4/3 - 0 = 4/3.

Solution: The integral of y=x2 + 1 for x between 0 and 1 is 4/3.

sigi 3 years ago

good stuff!