Evaluate the Definite Integrals
While evaluating definite integrals, we should need to calculate the closed area between the curve obtained by that function and the x-axis, over the given interval i.e. [a,b].
How to Solve Definite Integrals?
Consider we have a function f(x) on an interval [a,b], then the definite integral should be evaluated as follow:
$$ \int_a^b f(x) \; dx \;=\; F(b) \;-\; F(a) $$
Here is the basic rule for evaluating definite integrals. We have to solved first for the indefinite integral of that function and then it will be simple to find definite integral of a given function.
After finding the indefinite integral, we have to solved for upper and lower limits of that function as given in above formula. In this way you will be able to learn how to solve definite integrals step by step
Keeping the properties of definite integral and that formula for solving definite integral in mind, Let's move toward definite integral examples and practice them to evaluate the definite integrals.
To start out we will just integrate simple polynomials. This involves taking the integral of function.
Definite Integral Examples
- Consider we have a function
$$ \int_0^2 (x^2+1) \; dx $$
Let’s try to evaluate the definite integral given in example.
Solution:
$$ \int_0^2 (x^2+1) \; dx $$
Taking the anti-derivative, it will become
$$ \left( \frac{x^3}{3} + x \right) \Biggl|_0^2 $$ $$ F(2)-F(0) \;=\; \left( \frac{(2)^3}{3} + 2 \right) -\left( 0 \right) $$ $$ F(2)-F(0) \;=\; \frac{14}{3} - 0 $$ $$ \int_0^2 (x^2+1) \; dx \;=\; \frac{14}{3} $$
That's quite simple. Now Let's try another one to evaluate definite integral
- Consider we have a function
$$ \int_1^2 (x^3 + 2x + \frac{1}{x^2} ) \; dx $$
Let’s try to integrate that simple definite integral
Solution:
$$ \int_1^2 (x^3 + 2x + x^{-2} ) \; dx $$ $$ \implies \left( \frac{x^4}{4} + \frac{x^2}{2} - \frac{1}{x} \right) \Biggl|_1^2 $$ $$ F(2)-F(1) \;=\; \left( 4 + 4 - \frac{1}{2} \right) - \left( \frac{1}{4} + 1 - 1 \right) $$ $$ \int_1^2 (x^3 + 2x + \frac{1}{x^2} ) \; dx \;=\; 0 $$
- Consider we have a function
$$ \int_0^{\frac{π}{2}} cosx \; dx$$
Let’s try to integrate that trigonometric definite integral
Solution:
$$ \int_0^{\frac{π}{2}} cosx \; dx$$ $$ \implies \Biggr| sin x \Biggr|_0^{\frac{π}{2}} $$ $$ F(\frac{π}{2}) \;-\; F(0) \;=\; sin(\frac{π}{2}) \;-\; sin(0) $$ $$ F(\frac{π}{2}) \;-\; F(0) \;=\; 1 \;-\; 0 $$ $$ \int_0^{\frac{π}{2}} cosx \; dx \;=\; 1$$
So these are some definite integral examples that will be very helpful because any concept of mathematics is incomplete without examples.
Conclusion
Hopefully after computing some simple integrals we can see the immense power of this method. Prior to the fundamental theorem of calculus, finding areas and distances associated with curvature was incredibly complex, and only brilliant mathematicians could figure out how to do it.
$$ \int_a^b f(x) \; dx = F(b)-F(a) $$
With this simple algorithm of finding the antiderivative, anyone can answer how to solve definite integrals easily without any hurdle.