Question

Evaluate the following: \((a) \int_{1}^{3} \frac{1}{2} x^{2} d x\) \((d) \int_{2}^{4}\left(x^{3}-6 x^{2}\right) d x\) \((b) \int_{0}^{1} x\left(x^{2}+6\right) d x\) \((e) \int_{-1}^{1}\left(a x^{2}+b x+c\right) d x\) \((c) \int_{1}^{3} 3 \sqrt{x} d x\) \((f) \int_{4}^{2} x^{2}\left(\frac{1}{3} x^{3}+1\right) d x\)

Short answer

(a) \(\frac{13}{3}\); (d) -52; (b) \(\frac{13}{4}\); (e) \(2c\); (c) \(6\sqrt{3} - 2\); (f) -109.33

Step-by-step solution

Evaluating Part (a)

To evaluate the integral \( \int_{1}^{3} \frac{1}{2} x^{2} \, dx \), first find the antiderivative of \( \frac{1}{2} x^{2} \). This is \( \frac{1}{2} \cdot \frac{x^{3}}{3} = \frac{x^{3}}{6} \). Now evaluate the definite integral by calculating: \[ \left[ \frac{x^{3}}{6} \right]_{1}^{3} = \frac{3^{3}}{6} - \frac{1^{3}}{6} = \frac{27}{6} - \frac{1}{6} = \frac{26}{6} = \frac{13}{3}. \]

Evaluating Part (d)

First, split the integral: \( \int_{2}^{4}(x^{3} - 6x^{2}) \, dx \) into two parts: \( \int_{2}^{4} x^{3} \, dx - 6 \int_{2}^{4} x^{2} \, dx \). The antiderivatives are \( \frac{x^{4}}{4} \) and \( \frac{x^{3}}{3} \), respectively. Evaluate each one: \[ \left[ \frac{x^{4}}{4} \right]_{2}^{4} = \frac{4^{4}}{4} - \frac{2^{4}}{4} = 64 - 4 = 60 \] and \[ 6 \left[ \frac{x^{3}}{3} \right]_{2}^{4} = 6 \left( \frac{64}{3} - \frac{8}{3} \right) = 6 \cdot \frac{56}{3} = 6 \cdot \frac{56}{3} = \frac{336}{3} = 112. \] Subtract to find the integral: \( 60 - 112 = -52 \).

Evaluating Part (b)

Expand \( x(x^{2} + 6) \) to get \( x^{3} + 6x \). The antiderivatives are \( \frac{x^{4}}{4} \) and \( 3x^{2} \), respectively. Evaluate from 0 to 1: \[ \left[ \frac{x^{4}}{4} \right]_{0}^{1} + \left[ 3x^{2} \right]_{0}^{1} = \left( \frac{1}{4} - 0 \right) + \left( 3 \times 1^{2} - 3 \times 0^{2} \right) = \frac{1}{4} + 3 = \frac{13}{4}. \]

Evaluating Part (e)

The given polynomial is \( ax^{2} + bx + c \). Use the integral \( \int_{-1}^{1} ax^{2} \, dx + \int_{-1}^{1} bx \, dx + \int_{-1}^{1} c \, dx \). \( ax^{2} \) leads to \( \frac{ax^{3}}{3} \), which is zero on symmetric bounds. \( bx \) results in \( \frac{bx^{2}}{2} \), also zero when computed from \(-1\) to \(1\). Finally, \( c \int_{-1}^{1} \, dx = c[1 - (-1)] = 2c \). So the result is \( 2c \).

Evaluating Part (c)

Rewrite \( 3 \sqrt{x} \) as \( 3x^{1/2} \). The antiderivative is \( \frac{3x^{3/2}}{3/2} = 2x^{3/2} \). Evaluate from 1 to 3: \[ \left[ 2x^{3/2} \right]_{1}^{3} = \left( 2(3^{3/2}) - 2(1^{3/2}) \right) = 2(\sqrt{27}) - 2 = 6\sqrt{3} - 2. \]

Evaluating Part (f)

First distribute: \( x^{2}(\frac{1}{3}x^{3} + 1) = \frac{1}{3}x^{5} + x^{2} \). The antiderivatives are \( \frac{x^{6}}{18} \) and \( \frac{x^{3}}{3} \). Reverse the limits to evaluate: \[ \left[ \frac{x^{6}}{18} \right]_{4}^{2} + \left[ \frac{x^{3}}{3} \right]_{4}^{2} = \left( \frac{64}{3} - \frac{4096}{18} \right) - \left( \frac{8}{3} - \frac{64}{3} \right). \] Calculate each term: \( \frac{64}{18} - \frac{4096}{18} + \frac{8}{3} - \frac{64}{3} \). Simply: -109.33.

Key concepts

Antiderivative

When working with integrals, one of the fundamental tasks is to find the antiderivative of a given function. The antiderivative, often referred to as the indefinite integral, is essentially the reverse process of differentiation. If you differentiate the antiderivative of a function, you should get the original function back. For example, if the original function is \( \frac{1}{2} x^2 \), its antiderivative, as seen in Step 1 of the original solution, would be \( \frac{x^3}{6} \). This is because differentiating \( \frac{x^3}{6} \) with respect to \( x \) yields back \( \frac{1}{2} x^2 \).
To find an antiderivative, integrate each term of the polynomial by adding 1 to the exponent of \( x \) and then dividing by the new exponent. For constant terms, such as \( c \) in an expression \( a x^2 + b x + c \), the antiderivative is simply \( cx \). The introduction of the constant \( C \) (arbitrary constant of integration) in indefinite integrals is crucial but does not apply to definite integrals, which we deal with in this exercise.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is the cornerstone of calculus that connects the concepts of differentiation and integration. It has two parts, but in this context, the second part is most relevant, which allows us to evaluate definite integrals. The theorem states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \([a, b] \), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a).\]This means, to solve a definite integral, you can substitute the upper and lower limits of the integral into the antiderivative and subtract the latter from the former.
For instance, in Step 1 of the solution, after finding the antiderivative \( \frac{x^3}{6} \), you calculate \( \left[ \frac{x^3}{6} \right]_{1}^{3} = \frac{3^3}{6} - \frac{1^3}{6} \). This simple operation directly and elegantly gives the area under the curve from \( x = 1 \) to \( x = 3 \).

Polynomial Integration

Polynomial integration involves integrating terms that consist of variables raised to various powers, which is often straightforward due to its structured form. Each term \( ax^n \) of a polynomial is integrated using the power rule of integration.
Here is how you apply polynomial integration:

• Increase the exponent by one, so \( n \) becomes \( n+1 \).
• Divide the term by the new exponent \( n+1 \).

In practice, for Step 2 of the solution, \( x^3 \) becomes \( \frac{x^4}{4} \), and \( x^2 \) becomes \( \frac{x^3}{3} \). Remember to treat coefficients as constant multipliers, keeping them outside of the integral process until evaluating the result. Through polynomial integration, the computation simplifies complex problems into easier, manageable parts, ready for evaluation under defined limits.

Integral Evaluation Steps

Solving definite integrals involves a series of methodical steps, integrating the function across specific limits to find the area under the curve. Here’s a simple guide based on the examples provided:

• First, find the antiderivative of the given function or expressions, term by term.
• Apply the Fundamental Theorem of Calculus by evaluating this antiderivative at the given limits. Substitute the upper and lower limits into the antiderivative expression, then subtract the lower from the upper limit.
• If dealing with complex expressions or multiple terms, split the integral as needed, compute each part, and combine the results appropriately.
• A negative bound, such as seen in Step 6, requires reversing the integral limits and negating the result, simplifying calculations and ensuring correctness.

This step-by-step approach assures accurate results while enhancing your understanding of calculating areas, crucial for grasping integral calculus.