Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{1}(x+\sqrt{x}) d x$$

Short answer

Question: Evaluate the definite integral $\int_{0}^{1}(x+\sqrt{x}) dx$ using the Fundamental Theorem of Calculus. Answer: $\frac{5}{6}$

Step-by-step solution

Find the Antiderivative of the Function

We need to find the antiderivative of the function given inside the integral, which is \((x+\sqrt{x})\). To do this, we should rewrite the square root as a power and find the antiderivative of each term separately: $$x+\sqrt{x} = x + x^{\frac{1}{2}}$$ Now, find the antiderivative of each term: $$\int (x + x^{\frac{1}{2}})dx = \int x dx + \int x^{\frac{1}{2}} dx$$ Using the power rule for antiderivatives: $$\int x^a dx = \frac{x^{a+1}}{a+1} + C$$ We get: $$\frac{x^2}{2} + \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{x^2}{2} + \frac{2}{3}x^{\frac{3}{2}} + C$$

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$ Where \(F(x)\) is the antiderivative of \(f(x)\). Using the antiderivative we found in Step 1: $$F(x) = \frac{x^2}{2} + \frac{2}{3}x^{\frac{3}{2}} + C$$ We can now evaluate the definite integral: $$\int_{0}^{1}(x+\sqrt{x}) dx = F(1) - F(0)$$ $$= \left(\frac{1^2}{2} + \frac{2}{3}1^{\frac{3}{2}}\right) - \left(\frac{0^2}{2} + \frac{2}{3}0^{\frac{3}{2}}\right)$$ $$= \frac{1}{2} + \frac{2}{3} - 0$$ $$= \frac{5}{6}$$ So, the value of the definite integral is \(\frac{5}{6}\).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is one of the cornerstones of calculus. It states a profound relationship between differentiation and integration. This theorem links the concept of the derivative of a function to the concept of the integral.

• The first part of the theorem tells us that if we have a continuous function, the integral of the function over an interval can be found using its antiderivative.
• The second part allows us to evaluate the definite integral of a function using the antiderivatives.

The formula for the theorem involves evaluating the antiderivative at the bounds of the interval. This means if you know the antiderivative of a function, you can find the accumulation of quantities over that interval easily. The formula is written as:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]In our exercise, we apply the theorem to find the area under the curve of the function \((x + \sqrt{x})\) from 0 to 1.

Antiderivative

An antiderivative of a function is essentially the reverse process of differentiation. If you differentiate an antiderivative, you end up with your original function. Finding the antiderivative is a crucial step in solving definite integrals.For the given exercise, the original function we need the antiderivative for is \((x + \sqrt{x})\). We rewrite the square root term as a power, like this:

• \(x\) can be written as \(x^1\)
• \(\sqrt{x}\) is rewritten as \(x^{\frac{1}{2}}\)

Then, to find the antiderivative of these, we apply integration to each term separately, combining results. This step ensures each term's basis is communicated effectively, such as for \(x\), resulting in \(\frac{x^2}{2}\), and for \(x^{\frac{1}{2}}\), resulting in \(\frac{2}{3}x^{\frac{3}{2}}\). The antiderivative includes a constant \(C\), termed the constant of integration.

Power Rule

The power rule for integration is an easy handy tool for finding antiderivatives of power functions, which are functions of the form \(x^n\). The rule states that the integral of \(x^n\) with respect to \(x\) is:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \(n eq -1\), to ensure the denominator isn't zero. This rule applies not only to whole numbers but also to fractional and negative powers, making it versatile.In our example, we applied the power rule separately to \(x\) and \(x^{\frac{1}{2}}\):

• For \(x\), \(n = 1\), resulting in \(\frac{x^2}{2}\)
• For \(x^{\frac{1}{2}}\), \(n = \frac{1}{2}\), leading to \(\frac{2}{3}x^{\frac{3}{2}}\)

The power rule makes finding antiderivatives a straightforward process when dealing with polynomial terms.

Evaluation of Definite Integrals

The evaluation of definite integrals involves calculating the total accumulation between two points, often visualized as the area under a curve. The task is achieved by plugging the limits of integration, usually given as points \(a\) and \(b\), into the antiderivative. In our exercise, once we have the antiderivative \(F(x)\), we compute:

• \(F(1)\), the value at the upper limit
• \(F(0)\), the value at the lower limit

Then, apply the formula:\[ \int_{0}^{1} (x + \sqrt{x}) \, dx = F(1) - F(0) \]This leads to calculating actual values:

• \(F(1) = \frac{1}{2} + \frac{2}{3}\)
• Subtract \(F(0) = 0\)

The result \(\frac{5}{6}\) gives the exact area or "net change" between \(x = 0\) and \(x = 1\). Definite integrals provide precise numerical values, unlike indefinite integrals which leave results in terms of functions with a constant \(C\).