Problem 103 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{2} t(t+3) d t $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{26}{3} \).

Step 1: Expand the Integrand

First, distribute the terms within the integrand. We have the expression \( t(t+3) \). Expand this to get: \[ t(t+3) = t^2 + 3t \]

Step 2: Set Up the Integral

After expanding the integrand, we rewrite the definite integral with the expanded expression: \[ \int_{0}^{2} (t^2 + 3t) \, dt \]

Step 3: Integrate Each Term

Now, integrate each term individually. The integral of \( t^2 \) is \( \frac{t^3}{3} \) and the integral of \( 3t \) is \( \frac{3t^2}{2} \). This gives us: \[ \int (t^2 + 3t) \, dt = \frac{t^3}{3} + \frac{3t^2}{2} + C \] (Since this is a definite integral, we don't need the constant \( C \).)

Step 4: Evaluate the Definite Integral

Now, evaluate the antiderivative from the limits of 0 to 2: \[ \left[ \frac{t^3}{3} + \frac{3t^2}{2} \right]_{0}^{2} \]Calculate at the upper limit (2):\[ \left( \frac{2^3}{3} + \frac{3(2)^2}{2} \right) = \frac{8}{3} + \frac{12}{2} = \frac{8}{3} + 6 \]Then, calculate at the lower limit (0):\[ \left( \frac{0^3}{3} + \frac{3(0)^2}{2} \right) = 0 \]

Step 5: Compute the Final Result

Subtract the result at the lower limit from the result at the upper limit:\[ \left( \frac{8}{3} + 6 \right) - 0 = \frac{8}{3} + 6 \]Convert 6 to a fraction with a denominator of 3:\[ 6 = \frac{18}{3} \]Now add the fractions together:\[ \frac{8}{3} + \frac{18}{3} = \frac{26}{3} \]

Step 6: Conclusion

The value of the definite integral \( \int_{0}^{2} t(t+3) \, dt \) is \( \frac{26}{3} \).

Key Concepts

Expand Integrand

Before diving into complex processes like integration, it's essential to simplify our expression wherever possible. In this case, we start with the integrand \( t(t+3) \). This expression isn't in the friendliest form for integration. Expanding it gives a clearer picture of what we're dealing with and transforms it to something more manageable.
Expanding \( t(t+3) \) involves distributing each term to simplify the expression. We multiply \( t \) by each term inside the parenthesis:

• \( t \times t = t^2 \)
• \( t \times 3 = 3t \)

This results in the expanded form: \( t^2 + 3t \). By expanding, the integrand is now written in a polynomial form. This step is crucial because it's much easier to integrate when each term is clearly separated and simplified.

Integrate Term by Term

With the integrand now expanded to \( t^2 + 3t \), the next logical step is to integrate each term of this polynomial separately. This approach is rooted in one of the fundamental properties of integrals, which allows us to integrate term by term.
To integrate term by term means handling each component of our expression separately. Consider:

• The integral of \( t^2 \) is calculated by increasing the exponent by one and dividing by this new value. Thus, \( \int t^2 \, dt = \frac{t^3}{3} \).
• Similarly, \( \int 3t \, dt = \frac{3t^2}{2} \). Here we increase the exponent of \( t \) to 2, and divide by the new exponent.

After integrating each term, we compile these results to form our indefinite integral: \( \frac{t^3}{3} + \frac{3t^2}{2} \). Remember, when working with definite integrals, the constant of integration \( C \) can be omitted.

Evaluate Antiderivative

Once you've found the antiderivative, the next step is evaluating this expression from the lower limit to the upper limit of the definite integral, which transforms our solution from an indefinite to a definite form.
Specifically for \( \int_{0}^{2} (t^2 + 3t) \, dt \), we have calculated the antiderivative as \( \frac{t^3}{3} + \frac{3t^2}{2} \). To evaluate this, substitute the upper and lower bounds into the antiderivative:

• For the upper limit (2): \[ \left( \frac{2^3}{3} + \frac{3(2)^2}{2} \right) = \frac{8}{3} + 6 \]
• For the lower limit (0): \[ \left( \frac{0^3}{3} + \frac{3(0)^2}{2} \right) = 0 \]

Subtracting these results gives us the change in the function's value over our specified interval.

Compute Limits

The final act of finding a definite integral is to compute the limits by subtracting the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit. This step ensures that we consider the area under the curve between these two bounds.
From our previous calculations, we have:

• Upper limit (2): \( \frac{8}{3} + 6 \)
• Lower limit (0): \( 0 \)

To find the definite integral's value, perform the subtraction: \( \left( \frac{8}{3} + 6 \right) - 0 = \frac{8}{3} + 6 \).
To finalize, express 6 as a fraction with a common denominator of 3, \( 6 = \frac{18}{3} \). Adding these fractions together gives us \( \frac{26}{3} \) as the solution. This results in the final value of the definite integral, capturing the total area under the curve from \( t=0 \) to \( t=2 \).