Question text Find [math: ∫6x5+2x2−4x+5]\displaystyle\int{6\,x^5+2\,x^2-4\,x+5} [math: dx]. Note: Type "[math: c]" for the integral constant. [input] Check Question 15

Question

BlackTom AI · Accepted Answer

The task is to compute the indefinite integral of the polynomial 6x^5 + 2x^2 - 4x + 5 with respect to x.
First, integrate each term separately: for ∫6x^5 dx, the power rule gives x^6 since ∫x^n dx = x^{n+1}/(n+1); multiplying by 6 yields x^6.
Next, ∫2x^2 dx equals 2 * (x^3/3) = (2/3)x^3 by applying the same rule to x^2.
Then, ∫(-4x) dx is -4 * (x^2/2) = -2x^2, again using the power rule.
Finally, ∫5 dx is 5x, since the integral of a constant is that constant times x.
Combine all terms and add the constant of integration: x^6 + (2/3)x^3 - 2x^2 + 5x + C.
The provided answer, written as [math: x^6 + (2/3)x^3 - 2x^2 + 5x + c], matches this result, differing only in the conventional notation for the constant of integration (C vs c).
This aligns with standard antiderivative rules and the linearity of the integral, since each term was integrated independently and then summed.

Question text Find [math: ∫6x5+2x2−4x+5]\displaystyle\int{6\,x^5+2\,x^2-4\,x+5} [math: dx]. Note: Type "[math: c]" for the integral constant. [input] Check Question 15

查看解析

登录即可查看完整答案

类似问题

Integrate ∫ ( 2 𝑥 − 1 ) 4 𝑑 𝑥

Question at position 11 find ∫23−xdx\int2^{3-x}dx−e3−xln⁡(2)+C- \frac{e^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C \frac{2^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C\frac{2^{3 - x}}{\ln(2)} + C−23−xln⁡(2)+C- \frac{2^{3 - x}}{\ln(2)} + C

Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+Cx44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+C

Question at position 4 ∫2x−13dx=\int\frac{2x-1}{3}dx=(2x−1)26+C\frac{\left(2x-1\right)^2}{6}+C13(x2−x)+C\frac{1}{3}\left(x^2-x\right)+C12+C\frac{1}{2}+Cx2−x3x+C\frac{x^2-x}{3x}+C23+C\frac{2}{3}+C

Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+Cx44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+C

Question at position 5 ∫323dx=\int3^{\frac{2}{3}}dx=0+C0+C323x+C3^{\frac{2}{3}}x+C35353+C\frac{3}{5}3^{\frac{5}{3}}+C0053353+C\frac{5}{3}3^{\frac{5}{3}}+C

Question at position 4 ∫2x−13dx=\int\frac{2x-1}{3}dx=23+C\frac{2}{3}+C12+C\frac{1}{2}+C13(x2−x)+C\frac{1}{3}\left(x^2-x\right)+C(2x−1)26+C\frac{\left(2x-1\right)^2}{6}+Cx2−x3x+C\frac{x^2-x}{3x}+C

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单