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Question: How do you use the direct comparison test for improper integrals?...

How do you use the direct comparison test for improper integrals?

Explanation

Solution

Hint : Let’s take a function suppose f(x)f\left( x \right) and the function is greater than zero i.e. f(x)0f\left( x \right) \geqslant 0 with interval [a,)[a,\infty ) .Take any one example either for convergence or divergence to test the following. Suppose if a function is convergent then let’s take another function that is under the previous function and its also greater than zero, that means this function will automatically also converge. Similarly, check for the divergence.

Complete step by step solution:
Let’s take a function f(x)f\left( x \right) which is greater than or equal to zero i.e. f(x)0f\left( x \right) \geqslant 0 for interval [a,)[a,\infty ) and suppose that this function is convergent or converges.
If the function is convergent that means the area is finite.
Then take another function g(x)g\left( x \right) which is also greater than or equal to zero i.e. g(x)0g\left( x \right) \geqslant 0. But the function g(x)g\left( x \right) is under or inside the function f(x)f\left( x \right) .
That gives an Equation 0g(x)f(x)0 \leqslant g\left( x \right) \leqslant f\left( x \right) .
From the Equation and comparison, its absolutely clear that if the function f(x)f\left( x \right) converges (having a finite area) that means the function g(x)g\left( x \right) inside it will also have a finite area and it will also converge for the interval [a,)[a,\infty ) .
Therefore, this means that
If af(x)\int\limits_a^\infty {f\left( x \right)} converges then so does ag(x)\int\limits_a^\infty {g\left( x \right)} .
This is how we can use D.C.T or the direct comparison test for improper integrals.

Note : We can also take the example of divergence, we can also test that, but for divergence the area would be infinite for the interval given and the equation would be like 0f(x)g(x)0 \leqslant f\left( x \right) \leqslant g\left( x \right) .And, we would come for the same type of result that is: If af(x)\int\limits_a^\infty {f\left( x \right)} diverges then so does ag(x)\int\limits_a^\infty {g\left( x \right)} .