int_(0)^(1)(a+(b-a)t)^(k)dt,k inN.

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of the function (a + (b - a)t)^k with respect to t from 0 to 1, where k is a natural number.Apply Substitution Method: Apply the substitution method to simplify the integral. Let u = a + (b - a)t, which implies that du = (b - a)dt. We also need to change the limits of integration according to the substitution. When t = 0, u = a. When t = 1, u = b.Express dt in terms of du: Express dt in terms of du and adjust the integral accordingly. From du = (b - a)dt, we get dt = du / (b - a). Now we can rewrite the integral in terms of u: ∫(a + (b - a)t)^k dt from 0 to 1 = ∫u^k * (1/(b - a)) du from a to b.Evaluate Integral: Evaluate the integral with the new variable and limits. The integral becomes (1/(b - a)) * ∫u^k du from a to b. The antiderivative of u^k is u^(k+1)/(k+1), so we have: (1/(b - a)) * [u^(k+1)/(k+1)] from a to b.Substitute Limits: Substitute the limits into the antiderivative and calculate the difference. Substitute u = b and u = a into the antiderivative and calculate the difference: (1/(b - a)) * [b^(k+1)/(k+1) - a^(k+1)/(k+1)].Simplify Final Answer: Simplify the expression to get the final answer. The final answer is: (1/(b - a)) * (1/(k+1)) * [b^(k+1) - a^(k+1)].

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Calculus

Evaluate definite integrals using the chain rule

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\int_{0}^{1}(a+(b-a) t)^{k} d t, k \in \mathbb{N}

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $(a + (b - a)t)^k$ with respect to $t$ from $0$ to $1$ , where $k$ is a natural number.
Apply Substitution Method: Apply the substitution method to simplify the integral. Let $u = a + (b - a)t$ , which implies that $du = (b - a)dt$ . We also need to change the limits of integration according to the substitution. When $t = 0$ , $u = a$ . When $t = 1$ , $u = b$ .
Express $dt$ in terms of $du$ : Express $dt$ in terms of $du$ and adjust the integral accordingly. $\newline$ From $du = (b - a)dt$ , we get $dt = \frac{du}{(b - a)}$ . Now we can rewrite the integral in terms of $u$ : $\newline$ $\int_{0}^{1}(a + (b - a)t)^k dt = \int_{a}^{b}u^k \cdot \frac{1}{(b - a)} du$ .
Evaluate Integral: Evaluate the integral with the new variable and limits. $\newline$ The integral becomes $(1/(b - a)) \cdot \int_a^b u^k \, du$ . The antiderivative of $u^k$ is $u^{(k+1)}/(k+1)$ , so we have: $\newline$ $(1/(b - a)) \cdot [u^{(k+1)}/(k+1)]$ from $a$ to $b$ .
Substitute Limits: Substitute the limits into the antiderivative and calculate the difference. $\newline$ Substitute $u = b$ and $u = a$ into the antiderivative and calculate the difference: $\newline$ $\frac{1}{(b - a)} \cdot \left[\frac{b^{(k+1)}}{(k+1)} - \frac{a^{(k+1)}}{(k+1)}\right]$ .
Simplify Final Answer: Simplify the expression to get the final answer. $\newline$ The final answer is: $\newline$ $\frac{1}{(b - a)} \cdot \frac{1}{(k+1)} \cdot [b^{(k+1)} - a^{(k+1)}]$ .