QUESTION IMAGE
Question
evaluate the following integral.\\(\int 8x (x^3 + 5) \\, dx\\)\\(\int 8x (x^3 + 5) \\, dx = \square\\) (type an exact answer.)
Step1: Expand the integrand
First, we expand the function inside the integral. Using the distributive property (also known as the distributive law of multiplication over addition), we multiply \(8x\) by each term inside the parentheses \((x^{3}+5)\). So, \(8x(x^{3}+5)=8x\times x^{3}+8x\times5\). When multiplying variables with exponents, we use the rule \(a^{m}\times a^{n}=a^{m + n}\), so \(x\times x^{3}=x^{1+3}=x^{4}\). And \(8x\times5 = 40x\). So the integrand becomes \(8x^{4}+40x\).
Step2: Integrate term - by - term
We use the power rule for integration, which states that \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (where \(n
eq - 1\) and \(C\) is the constant of integration).
- For the first term \(8x^{4}\):
Using the power rule, \(\int8x^{4}dx = 8\times\frac{x^{4 + 1}}{4+1}+C_{1}=8\times\frac{x^{5}}{5}+C_{1}=\frac{8}{5}x^{5}+C_{1}\)
- For the second term \(40x\):
Using the power rule, since \(x = x^{1}\), \(\int40x dx=40\times\frac{x^{1+1}}{1 + 1}+C_{2}=40\times\frac{x^{2}}{2}+C_{2}=20x^{2}+C_{2}\)
Step3: Combine the results
The integral of the sum of two functions is the sum of their integrals. So \(\int(8x^{4}+40x)dx=\int8x^{4}dx+\int40x dx\). Combining the two results from the previous step and combining the constants of integration \(C = C_{1}+C_{2}\), we get \(\frac{8}{5}x^{5}+20x^{2}+C\). Since the problem asks for an exact answer (and in the context of indefinite integrals, we can write the answer with the constant of integration, or if we assume \(C = 0\) for the purpose of the boxed answer as it's a general antiderivative), the answer is \(\frac{8}{5}x^{5}+20x^{2}+C\) (or \(\frac{8}{5}x^{5}+20x^{2}\) if we ignore the constant for the boxed form as the problem says "type an exact answer" and in some cases, the constant is implied or can be included as \(C\)).
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\(\frac{8}{5}x^{5}+20x^{2}+C\) (or \(\frac{8}{5}x^{5}+20x^{2}\) if \(C = 0\) is assumed)