Definite Integrals: Step-by-Step Evaluation Guide

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Hey guys! Today, we're diving deep into the world of definite integrals. If you've ever wondered how to calculate the area under a curve between two specific points, you're in the right place. We're going to break down five different definite integrals step by step, so you can master this crucial calculus concept. Let's get started!

1. Evaluating ∫ from 1 to 4 of x^4 dx

Our first challenge is to evaluate the definite integral of x^4 from 1 to 4. Remember, the fundamental theorem of calculus is our best friend here. It tells us that to evaluate a definite integral, we first need to find the antiderivative of the function and then evaluate it at the upper and lower limits of integration. Finally, we subtract the value at the lower limit from the value at the upper limit.

So, let’s break this down:

Finding the Antiderivative

The power rule for integration states that ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Applying this rule to x^4, we get:

∫x^4 dx = (x^(4+1))/(4+1) + C = (x^5)/5 + C

Evaluating at the Limits

Now, we evaluate the antiderivative at the upper limit (4) and the lower limit (1):

  • At x = 4: (4^5)/5 = 1024/5
  • At x = 1: (1^5)/5 = 1/5

Calculating the Definite Integral

Finally, we subtract the value at the lower limit from the value at the upper limit:

∫ from 1 to 4 of x^4 dx = (1024/5) - (1/5) = 1023/5 = 204.6

Therefore, the definite integral of x^4 from 1 to 4 is 204.6. See? That wasn't so bad, was it? This exercise highlights the direct application of the power rule and the fundamental theorem, which are cornerstones in evaluating definite integrals. Remember, guys, practice makes perfect!

2. Evaluating ∫ from 2 to 3 of (1/2)x dx

Next up, we're tackling the definite integral of (1/2)x from 2 to 3. This one is a bit simpler, but it's still a great way to reinforce the fundamentals. The constant multiple rule will come in handy here, allowing us to pull the (1/2) out of the integral, simplifying the process. Let's dive in!

Finding the Antiderivative

First, we find the antiderivative of (1/2)x. We can use the power rule again, but this time, we also have a constant multiplier:

∫(1/2)x dx = (1/2)∫x dx = (1/2) * (x^2/2) + C = x^2/4 + C

Evaluating at the Limits

Now, we evaluate the antiderivative at the upper limit (3) and the lower limit (2):

  • At x = 3: (3^2)/4 = 9/4
  • At x = 2: (2^2)/4 = 4/4 = 1

Calculating the Definite Integral

Finally, we subtract the value at the lower limit from the value at the upper limit:

∫ from 2 to 3 of (1/2)x dx = (9/4) - 1 = 9/4 - 4/4 = 5/4 = 1.25

So, the definite integral of (1/2)x from 2 to 3 is 1.25. This example underscores how constant multiples can be easily handled in integration and how a straightforward application of the power rule and the fundamental theorem leads to the solution. Always remember, guys, the basics are the building blocks for more complex problems!

3. Evaluating ∫ from 0 to 10 of (x^2 - x + 1) dx

Now, let's move on to something a little more complex: the definite integral of (x^2 - x + 1) from 0 to 10. This integral involves a polynomial, which means we'll need to apply the power rule to each term separately. But don't worry, we've got this! Remember, the key to integrating polynomials is to take it one term at a time. Let's break it down, shall we?

Finding the Antiderivative

We find the antiderivative of each term in the polynomial separately:

∫(x^2 - x + 1) dx = ∫x^2 dx - ∫x dx + ∫1 dx

Applying the power rule to each term, we get:

  • ∫x^2 dx = (x^3)/3 + C
  • ∫x dx = (x^2)/2 + C
  • ∫1 dx = x + C

Combining these, the antiderivative of (x^2 - x + 1) is:

(x^3)/3 - (x^2)/2 + x + C

Evaluating at the Limits

Next, we evaluate the antiderivative at the upper limit (10) and the lower limit (0):

  • At x = 10: (10^3)/3 - (10^2)/2 + 10 = 1000/3 - 100/2 + 10 = 1000/3 - 50 + 10 = 1000/3 - 40
  • At x = 0: (0^3)/3 - (0^2)/2 + 0 = 0

Calculating the Definite Integral

Finally, we subtract the value at the lower limit from the value at the upper limit:

∫ from 0 to 10 of (x^2 - x + 1) dx = (1000/3 - 40) - 0 = 1000/3 - 120/3 = 880/3 ≈ 293.33

Thus, the definite integral of (x^2 - x + 1) from 0 to 10 is approximately 293.33. This example illustrates how to integrate polynomials by applying the power rule term by term, highlighting the linearity of integration. Keep practicing, guys, and these polynomial integrals will become second nature!

4. Evaluating ∫ from 0 to 1 of (3x + 2)^2 dx

Alright, let's crank up the complexity a notch! Our next integral is ∫ from 0 to 1 of (3x + 2)^2 dx. This one involves a composite function raised to a power, so we'll need to think a little strategically. The most straightforward approach here is to first expand the square and then integrate term by term. This simplifies the problem into a polynomial integration, which we already know how to handle. Let's get to it!

Expanding the Square

First, we expand (3x + 2)^2:

(3x + 2)^2 = (3x + 2)(3x + 2) = 9x^2 + 12x + 4

Finding the Antiderivative

Now, we can integrate the expanded polynomial:

∫(9x^2 + 12x + 4) dx = ∫9x^2 dx + ∫12x dx + ∫4 dx

Applying the power rule to each term, we get:

  • ∫9x^2 dx = 9 * (x^3/3) = 3x^3 + C
  • ∫12x dx = 12 * (x^2/2) = 6x^2 + C
  • ∫4 dx = 4x + C

Combining these, the antiderivative of (3x + 2)^2 is:

3x^3 + 6x^2 + 4x + C

Evaluating at the Limits

Next, we evaluate the antiderivative at the upper limit (1) and the lower limit (0):

  • At x = 1: 3(1^3) + 6(1^2) + 4(1) = 3 + 6 + 4 = 13
  • At x = 0: 3(0^3) + 6(0^2) + 4(0) = 0

Calculating the Definite Integral

Finally, we subtract the value at the lower limit from the value at the upper limit:

∫ from 0 to 1 of (3x + 2)^2 dx = 13 - 0 = 13

Thus, the definite integral of (3x + 2)^2 from 0 to 1 is 13. This example emphasizes the importance of algebraic manipulation before integration, particularly expanding squares or other expressions to simplify the integrand. Great job, guys! You're crushing it!

5. Evaluating ∫ from 1 to 4 of dx/x

Last but definitely not least, we're tackling the definite integral of dx/x from 1 to 4. This one is special because it involves the reciprocal function, and its integral is a logarithm. This is a classic integral that every calculus student should know. So, let's finish strong and see how it's done!

Finding the Antiderivative

Remember, the antiderivative of 1/x is the natural logarithm of the absolute value of x:

∫(1/x) dx = ln|x| + C

Evaluating at the Limits

Now, we evaluate the antiderivative at the upper limit (4) and the lower limit (1):

  • At x = 4: ln|4| = ln(4)
  • At x = 1: ln|1| = ln(1) = 0

Calculating the Definite Integral

Finally, we subtract the value at the lower limit from the value at the upper limit:

∫ from 1 to 4 of dx/x = ln(4) - 0 = ln(4)

So, the definite integral of dx/x from 1 to 4 is ln(4). This result highlights the importance of recognizing standard integrals, like the integral of 1/x, which appear frequently in calculus. Fantastic work, guys! You've made it to the end!

Conclusion

And there you have it, guys! We've successfully evaluated five different definite integrals, each with its own unique challenges and insights. From simple power rule applications to polynomial integration and logarithmic functions, we've covered a lot of ground today. The key takeaway here is that with a solid understanding of the fundamental theorem of calculus, the power rule, and a bit of algebraic manipulation, you can tackle a wide range of definite integrals. Keep practicing, keep exploring, and you'll become a definite integral master in no time!