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Evaluating mathematical expression
I’m looking at the equation -1 minus -2, which is straightforward. Then I compute -1 divided by 4 and check if subtracting negative one-third is needed. I’m also evaluating -x cubed divided by 3 at x = -1, which should equal positive one-third. I think that comes out to one-twelfth. I’m trying to make sure I'm not missing any steps or miscalculating anywhere along the way!Calculating equations and differences
I’m working through some equations, checking values for F at different points. First, I calculated the result for F at -4 and found it to be 34.667. Wait, though! I remember that I had computed it differently earlier—turns out it should be -128/3 + 120, which gives me 77.333 instead. So, I’ll need to correct that difference too. It’s a bit of a puzzle I’m figuring out here!Verifying integrals and values
I’m calculating integrals to ensure accuracy. For the second integral from -1 to 3, I find the antiderivative and evaluate it, confirming it matches up. Then, I assess the first integral, ensuring I get the correct outputs and antiderivatives. I also need to explain why the results are negative, which seems consistent due to the integrands’ properties. The user’s query about verifying results encourages me to retain the precise input format.Ensuring readable input
I realize providing fractions is preferable, so I should mention that if the system accepts them, I can enter values like 125/4, 123/4, and -1805/6. I also think a screenshot of the fields could clarify the answer for the user. It's good to consider all aspects to avoid any issues with the final input marker. I want to make sure all formulas are clear and easy to read!
Краткий ответ:
- а) \(\displaystyle \frac{125}{4}\)
- б) \(\displaystyle \frac{123}{4}\)
- в) \(\displaystyle -\frac{1805}{6}\)
Модуль раскрываем по точкам, где выражение под знаком модуля равно нулю.
а)
\[ 4x-5=0 \Rightarrow x=\frac54. \]
\[ \begin{aligned} \int_0^5 |4x-5|\,dx &=\int_0^{5/4}(5-4x)\,dx+\int_{5/4}^{5}(4x-5)\,dx\\ &=\frac{25}{8}+\frac{225}{8} =\frac{250}{8} =\boxed{\frac{125}{4}}. \end{aligned} \]
б)
\[ x+1=0 \Rightarrow x=-1. \]
\[ \begin{aligned} \int_{-2}^{3}x^2|x+1|\,dx &=\int_{-2}^{-1}x^2(-x-1)\,dx +\int_{-1}^{3}x^2(x+1)\,dx\\ &=\frac{17}{12}+\frac{88}{3}\\ &=\frac{17}{12}+\frac{352}{12} =\boxed{\frac{123}{4}}. \end{aligned} \]
в)
На отрезке \([-7;0]\):
\[ 3x+12=0 \Rightarrow x=-4, \qquad |x-3|=3-x. \]
Поэтому
\[ \begin{aligned} \int_{-7}^{0}x\bigl(|3x+12|+|x-3|\bigr)\,dx ={}&\int_{-7}^{-4}x(-4x-9)\,dx\\ &+\int_{-4}^{0}x(2x+15)\,dx\\ ={}&-\frac{447}{2}-\frac{232}{3}\\ ={}&\boxed{-\frac{1805}{6}}. \end{aligned} \]
Проверка: в пунктах а) и б) подынтегральные функции неотрицательны, поэтому ответы положительные. В пункте в) \(x\le 0\), а сумма модулей неотрицательна, поэтому результат должен быть отрицательным.
