Solving Definite Integrals: A Step-by-Step Guide

Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of calculus, specifically focusing on how to evaluate definite integrals. Definite integrals are powerful tools that allow us to calculate the exact area under a curve between two specific points. This is incredibly useful in various fields, from physics and engineering to economics and statistics. We'll be tackling a specific example: $\int_3^4 \frac{t^5-3 t^2}{t^4} d t$. Don't let the notation intimidate you; by breaking it down, we can conquer it with ease. Understanding definite integrals is a fundamental skill for anyone serious about mathematics or its applications, and our journey today will provide you with a clear, actionable approach to solving such problems. We'll explore the foundational concepts, the practical steps involved in the evaluation, and the underlying principles that make this mathematical operation so significant. So, grab your thinking caps, and let's embark on this mathematical exploration together!

Understanding the Definite Integral

Before we jump into solving our specific problem, let's take a moment to truly grasp what a definite integral represents. At its core, a definite integral, denoted by $\int_a^b f(t) dt$, is a mathematical concept that measures the net area between a function's graph, the t-axis, and the vertical lines $t=a$ and $t=b$. Think of it as summing up an infinite number of infinitesimally small slices of area. The function $f(t)$ is the integrand, which describes the curve we're interested in. The limits of integration, $a$ (the lower limit) and $b$ (the upper limit), define the interval over which we are calculating this area. The '$dt$' signifies that we are integrating with respect to the variable $t$. The Fundamental Theorem of Calculus is the cornerstone that connects differentiation and integration, providing us with a method to evaluate definite integrals. It states that if $F(t)$ is an antiderivative of $f(t)$ (meaning $F'(t) = f(t)$), then $\int_a^b f(t) dt = F(b) - F(a)$. This theorem transforms the problem of summing infinitely many rectangles into finding an antiderivative and evaluating it at the limits. This is a monumental simplification, and it's the key to solving most definite integral problems. Without it, we would be stuck with the much more complex process of Riemann sums. Therefore, when we approach a definite integral, our primary goal is often to find the antiderivative of the integrand.

Simplifying the Integrand

Our definite integral is $\int_3^4 \frac{t^5-3 t^2}{t^4} d t$. The first crucial step in evaluating this integral is to simplify the integrand. The expression $\frac{t^5-3 t^2}{t^4}$ looks a bit cumbersome as it is. However, we can break it down into simpler terms by dividing each term in the numerator by the denominator. This is a fundamental algebraic manipulation that will make the integration process much more manageable. So, let's rewrite the integrand:

$\frac{t^5-3 t^2}{t^4} = \frac{t^5}{t^4} - \frac{3 t^2}{t^4}$

Now, we can use the rules of exponents (specifically, $t^m / t^n = t^{m-n}$) to simplify each of these fractions:

$\frac{t^5}{t^4} = t^{5-4} = t^1 = t$

and

$\frac{3 t^2}{t^4} = 3 t^{2-4} = 3 t^{-2}$

So, our simplified integrand becomes:

$t - 3 t^{-2}$

This simplified form is much easier to work with when finding the antiderivative. This algebraic simplification is a common and essential technique in calculus. It allows us to transform complex-looking expressions into forms that we can readily integrate using standard power rules. Always look for opportunities to simplify your integrand before attempting to find its antiderivative. This initial step can save you a significant amount of time and reduce the chance of errors. It’s like preparing your ingredients before cooking; a little effort upfront makes the main process much smoother. We've successfully transformed our integral into $\int_3^4 (t - 3 t^{-2}) dt$. The next phase involves finding the antiderivative of this simplified expression.

Finding the Antiderivative

With our integrand simplified to $t - 3 t^{-2}$, the next logical step is to find the antiderivative. Remember, the antiderivative of a function $f(t)$ is a function $F(t)$ whose derivative is $f(t)$. We'll use the power rule for integration, which states that $\int t^n dt = \frac{t^{n+1}}{n+1} + C$, where $n \neq -1$, and $C$ is the constant of integration. We can integrate each term of our simplified expression separately because integration is a linear operation.

Let's start with the first term, $t$. Here, $n=1$. Applying the power rule:

$\int t dt = \frac{t^{1+1}}{1+1} = \frac{t^2}{2}$

Now, let's consider the second term, $-3 t^{-2}$. Here, we have a constant coefficient $-3$ and $n=-2$. Applying the power rule:

$\int -3 t^{-2} dt = -3 \int t^{-2} dt$

Using the power rule for $t^{-2}$:

$-3 \left( \frac{t^{-2+1}}{-2+1} \right) = -3 \left( \frac{t^{-1}}{-1} \right)$

Simplifying this gives us:

$-3 (-t^{-1}) = 3 t^{-1}$

Combining the antiderivatives of both terms, the antiderivative of $t - 3 t^{-2}$ is:

$F(t) = \frac{t^2}{2} + 3 t^{-1}$

We can also write $3t^{-1}$ as $\frac{3}{t}$. So, our antiderivative is:

$F(t) = \frac{t^2}{2} + \frac{3}{t}$

For definite integrals, we don't need to include the constant of integration, $C$, because it cancels out when we evaluate $F(b) - F(a)$. This process of finding the antiderivative is fundamental to calculus. It's the inverse operation of differentiation. Mastering the power rule and other integration techniques for various functions is crucial for successfully solving calculus problems. The simplification we did earlier made finding this antiderivative straightforward. If we had tried to find the antiderivative of the original fraction directly, it would have been a much more complex task, possibly requiring techniques like integration by parts or trigonometric substitution, which are not needed here thanks to our initial algebraic step.

Evaluating the Definite Integral

Now that we have successfully found the antiderivative, $F(t) = \frac{t^2}{2} + \frac{3}{t}$, we can proceed to evaluate the definite integral using the Fundamental Theorem of Calculus. This involves plugging in the upper limit ($b=4$) and the lower limit ($a=3$) into our antiderivative and then subtracting the result of the lower limit from the result of the upper limit. Mathematically, this is expressed as $F(b) - F(a)$.

Let's first evaluate $F(t)$ at the upper limit, $t=4$:

$F(4) = \frac{(4)^2}{2} + \frac{3}{4}$

$F(4) = \frac{16}{2} + \frac{3}{4}$

$F(4) = 8 + \frac{3}{4}$

To add these, we find a common denominator:

$F(4) = \frac{32}{4} + \frac{3}{4} = \frac{35}{4}$

Next, let's evaluate $F(t)$ at the lower limit, $t=3$:

$F(3) = \frac{(3)^2}{2} + \frac{3}{3}$

$F(3) = \frac{9}{2} + 1$

Again, finding a common denominator:

$F(3) = \frac{9}{2} + \frac{2}{2} = \frac{11}{2}$

Finally, we subtract $F(3)$ from $F(4)$:

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = F(4) - F(3)$

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = \frac{35}{4} - \frac{11}{2}$

To subtract these fractions, we need a common denominator, which is 4:

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = \frac{35}{4} - \frac{11 \times 2}{2 \times 2}$

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = \frac{35}{4} - \frac{22}{4}$

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = \frac{35 - 22}{4}$

$\int_3^4 \frac{t^5-3 t^2}{t^4} d t = \frac{13}{4}$

So, the value of the definite integral is $\frac{13}{4}$. This final step, the evaluation, is where we get our numerical answer. It's the culmination of all the previous work: simplifying the integrand, finding the antiderivative, and then applying the Fundamental Theorem of Calculus. Each step builds upon the last, and precision at each stage is key to arriving at the correct final result. The result, $\frac{13}{4}$, represents the net signed area under the curve of the function $\frac{t^5-3 t^2}{t^4}$ between $t=3$ and $t=4$.

Conclusion

We have successfully navigated the process of evaluating a definite integral, specifically $\int_3^4 \frac{t^5-3 t^2}{t^4} d t$. By breaking down the problem into manageable steps, we first simplified the integrand algebraically to $t - 3 t^{-2}$. Then, we applied the power rule to find its antiderivative, which turned out to be $\frac{t^2}{2} + \frac{3}{t}$. Finally, using the Fundamental Theorem of Calculus, we evaluated this antiderivative at the upper and lower limits of integration and subtracted the results, yielding a final value of $\frac{13}{4}$. This exercise highlights the importance of algebraic manipulation prior to integration and the systematic application of calculus rules. Remember, every calculus problem is an opportunity to strengthen your understanding and problem-solving skills. Keep practicing, and don't be afraid to revisit the fundamental concepts. The world of mathematics is full of exciting challenges and rewarding discoveries!

For further exploration into the fascinating realm of calculus and its applications, you might find the resources at Khan Academy to be incredibly helpful. They offer a vast array of free courses, exercises, and explanations covering topics from basic algebra to advanced calculus. Another excellent resource for mathematical learning and problem-solving is Brilliant.org, which provides interactive courses and challenges designed to build a deep understanding of STEM subjects. These platforms can offer additional perspectives and practice opportunities to solidify your grasp on definite integrals and other calculus concepts.