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Integrate the following functions using the appropriate rules of integration. a. \( \int_{1}^{2} x \ln (x) d x \) (5 Marks) b. \( \int_{0}^{2} x\left(x^{2}-1\right)^{7} d x \) (7 Marks)

Ask by Clark Welch. in South Africa
Mar 16,2025

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The integral of \( \int_{1}^{2} x \ln (x) d x \) is \( \frac{8\ln{(2)}-3}{4} \). The integral of \( \int_{0}^{2} x\left(x^{2}-1\right)^{7} d x \) is 410.

Solution

Calculate the integral \( \int_{1}^{2} x \ln (x) d x \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int_{1}^{2} x\ln{\left(x\right)} dx\) - step1: Evaluate the integral: \(\int x\ln{\left(x\right)} dx\) - step2: Prepare for integration by parts: \(\begin{align}&u=\ln{\left(x\right)}\\&dv=xdx\end{align}\) - step3: Calculate the derivative: \(\begin{align}&du=\frac{1}{x}dx\\&dv=xdx\end{align}\) - step4: Evaluate the integral: \(\begin{align}&du=\frac{1}{x}dx\\&v=\frac{x^{2}}{2}\end{align}\) - step5: Substitute the values into formula: \(\ln{\left(x\right)}\times \frac{x^{2}}{2}-\int \frac{1}{x}\times \frac{x^{2}}{2} dx\) - step6: Calculate: \(\frac{x^{2}\ln{\left(x\right)}}{2}-\int \frac{x}{2} dx\) - step7: Evaluate the integral: \(\frac{x^{2}\ln{\left(x\right)}}{2}-\frac{x^{2}}{4}\) - step8: Simplify the expression: \(\frac{1}{2}x^{2}\ln{\left(x\right)}-\frac{x^{2}}{4}\) - step9: Return the limits: \(\left(\frac{1}{2}x^{2}\ln{\left(x\right)}-\frac{x^{2}}{4}\right)\bigg |_{1}^{2}\) - step10: Calculate the value: \(2\ln{\left(2\right)}-\frac{3}{4}\) - step11: Calculate: \(\frac{8\ln{\left(2\right)}-3}{4}\) Calculate the integral \( \int_{0}^{2} x\left(x^{2}-1\right)^{7} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{2} x\left(x^{2}-1\right)^{7} dx\) - step1: Evaluate the integral: \(\int x\left(x^{2}-1\right)^{7} dx\) - step2: Use the substitution \(dx=\frac{1}{2x} dt\) to transform the integral\(:\) \(\int x\left(x^{2}-1\right)^{7}\times \frac{1}{2x} dt\) - step3: Simplify: \(\int \frac{\left(x^{2}-1\right)^{7}}{2} dt\) - step4: Use the substitution \(t=x^{2}\) to transform the integral\(:\) \(\int \frac{\left(t-1\right)^{7}}{2} dt\) - step5: Rewrite the expression: \(\int \frac{1}{2}\left(t-1\right)^{7} dt\) - step6: Use properties of integrals: \(\frac{1}{2}\times \int \left(t-1\right)^{7} dt\) - step7: Use the substitution \(dt=1 dv\) to transform the integral\(:\) \(\frac{1}{2}\times \int \left(t-1\right)^{7}\times 1 dv\) - step8: Simplify: \(\frac{1}{2}\times \int \left(t-1\right)^{7} dv\) - step9: Use the substitution \(v=t-1\) to transform the integral\(:\) \(\frac{1}{2}\times \int v^{7} dv\) - step10: Evaluate the integral: \(\frac{1}{2}\times \frac{v^{7+1}}{7+1}\) - step11: Simplify: \(\frac{1}{2}\times \frac{v^{8}}{8}\) - step12: Multiply the terms: \(\frac{v^{8}}{2\times 8}\) - step13: Multiply the terms: \(\frac{v^{8}}{16}\) - step14: Substitute back: \(\frac{\left(t-1\right)^{8}}{16}\) - step15: Substitute back: \(\frac{\left(x^{2}-1\right)^{8}}{16}\) - step16: Return the limits: \(\left(\frac{\left(x^{2}-1\right)^{8}}{16}\right)\bigg |_{0}^{2}\) - step17: Calculate the value: \(410\) The integral of \( \int_{1}^{2} x \ln (x) d x \) is \( \frac{8\ln{(2)}-3}{4} \). The integral of \( \int_{0}^{2} x\left(x^{2}-1\right)^{7} d x \) is 410.

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To solve the integrals, we'll proceed step by step. a. We want to calculate \( \int_{1}^{2} x \ln(x) \, dx \). To do this, we can use integration by parts, which states: \[ \int u \, dv = uv - \int v \, du \] Let \( u = \ln(x) \) (thus \( du = \frac{1}{x} \, dx \)) and \( dv = x \, dx \) (therefore \( v = \frac{x^2}{2} \)). Applying integration by parts: \[ \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx \] Now, evaluating the integral \( \int x \, dx = \frac{x^2}{2} \) yields: \[ \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \frac{1}{4} x^2 + C \] Now we need to evaluate from 1 to 2: \[ \left. \left( \frac{x^2}{2} \ln(x) - \frac{1}{4} x^2 \right) \right|_{1}^{2} \] Substituting the bounds gives us: At \( x = 2 \): \[ \frac{2^2}{2} \ln(2) - \frac{1}{4}(2^2) = 2 \ln(2) - 1 \] At \( x = 1 \): \[ \frac{1^2}{2} \ln(1) - \frac{1}{4}(1^2) = 0 - \frac{1}{4} = -\frac{1}{4} \] Thus, we combine these results: \[ \left( 2 \ln(2) - 1 \right) - \left( -\frac{1}{4} \right) = 2 \ln(2) - 1 + \frac{1}{4} = 2 \ln(2) - \frac{4}{4} + \frac{1}{4} = 2 \ln(2) - \frac{3}{4} \] So, the value of the integral is: \[ \int_{1}^{2} x \ln(x) \, dx = 2 \ln(2) - \frac{3}{4}. \] b. Now for the integral \( \int_{0}^{2} x (x^2 - 1)^7 \, dx \). Here, we can also use substitution. Let \( u = x^2 - 1 \). Then, \( du = 2x \, dx \), or \( dx = \frac{du}{2x} \), which means that \( x = \sqrt{u + 1} \). Now, adjusting the limits: When \( x = 0 \), \( u = 0^2 - 1 = -1 \). When \( x = 2 \), \( u = 2^2 - 1 = 3 \). Now substituting: \[ \int_{0}^{2} x (x^2 - 1)^7 \, dx = \frac{1}{2} \int_{-1}^{3} (u)^{7} \, du \] The integral of \( u^7 \) is: \[ \frac{u^8}{8} \] So we evaluate: \[ \frac{1}{2} \left[ \frac{u^8}{8} \right]_{-1}^{3} = \frac{1}{2} \left[ \frac{3^8}{8} - \frac{(-1)^8}{8} \right] = \frac{1}{2} \left[ \frac{6561}{8} - \frac{1}{8} \right] = \frac{1}{2} \cdot \frac{6560}{8} = \frac{6560}{16} = 410 \] Thus, the integral is: \[ \int_{0}^{2} x (x^2 - 1)^7 \, dx = 410. \]

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