Find the indefinite integral and check the result by differentiation.
∫((t^3/3)+(1/4t²)dt
To find the indefinite integral of the given function ∫((t^3/3) + (1/4t^2)) dt, you can use the power rule and the rule for the integral of a reciprocal function.
Let's break it down step by step:
Step 1: Integrate (t^3/3)
Using the power rule, the integral of t^n is (t^(n+1))/(n+1), where n is any real number except -1.
So, integrating (t^3/3) will give us ((1/3) * t^4).
Step 2: Integrate (1/4t^2)
The integral of (1/t^2) can be found using the rule for the integral of a reciprocal function. The integral of (1/t^2) is -1/t.
Therefore, integrating (1/4t^2) gives us ((1/4) * (-1/t)), which simplifies to (-1/4t).
Step 3: Add the two integrals
Putting the results from Step 1 and Step 2 together, the indefinite integral of ((t^3/3) + (1/4t^2)) dt is ((1/3) * t^4) + (-1/4t).
To check if the result is correct, we can differentiate the obtained expression and see if it matches the original function.
Step 4: Differentiate the expression
Differentiating ((1/3) * t^4) + (-1/4t) with respect to t will give us (4/3 * t^3) + (1/4t^2).
Step 5: Compare with the original expression
If the result from Step 4 matches the original function ((t^3/3) + (1/4t^2)), then our solution is correct.
So, by differentiating the expression obtained in Step 4, we get (4/3 * t^3) + (1/4t^2), which is equal to the original function. Hence, our result is correct.
Therefore, the indefinite integral of ((t^3/3) + (1/4t^2)) dt is ((1/3) * t^4) + (-1/4t), and this result is confirmed by its differentiation.