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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?
- 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.
How to differentiate a power function, xn
- Multiply by the original exponent (n)
- 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...)
- Increase the exponent (n) by one (to give n+1)
- 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
- 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 = x2 + 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 = ∫ (x2 + 1) dx = x3 / 3 + x
- At x = 1, x3 / 3 + x = 4/3. At x = 0, x3 / 3 + x = 0
- 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.