微积分/微积分基本定理/解答

1. 计算 $\int _{0}^{1}x^{6}dx$。将你的答案与你在第 4.1 节练习 1 中得到的答案进行比较。

$\int _{0}^{1}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{0}^{1}={\frac {1^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {1}{7}}=0.{\overline {142857}}}$
这与我们在第 4.1 节练习 1 中计算出的边界一致。

2. 计算 $\int _{1}^{2}x^{6}dx$。将你的答案与你在第 4.1 节练习 2 中得到的答案进行比较。

$\int _{1}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{1}^{2}={\frac {2^{7}}{7}}-{\frac {1^{7}}{7}}={\frac {128}{7}}-{\frac {1}{7}}=\mathbf {{\frac {127}{7}}=18.{\overline {142857}}}$
这与我们在第 4.1 节练习 2 中计算出的边界一致。

3. 计算 $\int _{0}^{2}x^{6}dx$。将你的答案与你在第 4.1 节练习 4 中得到的答案进行比较。

$\int _{0}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{0}^{2}={\frac {2^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {128}{7}}=18.{\overline {285714}}}$
这与我们在第 4.1 节练习 4 中计算出的边界一致。

4. 计算 $\int \limits _{2}^{3}{\frac {x^{3}+3x^{2}-4}{x^{2}}}dx$.

5. 计算 $\int \limits _{1}^{4}2x^{3}+{\frac {1}{3x^{2}}}-4{\sqrt {x}}+5dx$.$${\begin{aligned}\int \limits _{1}^{4}2x^{3}+{\frac {1}{3x^{2}}}-4{\sqrt {x}}+5dx&=\int \limits _{1}^{4}2x^{3}dx+\int \limits _{1}^{4}{\frac {1}{3x^{2}}}dx-\int \limits _{1}^{4}4{\sqrt {x}}dx+\int \limits _{1}^{4}5dx\\&=2\int \limits _{1}^{4}x^{3}dx+{\frac {1}{3}}\int \limits _{1}^{4}{\frac {1}{x^{2}}}dx-4\int \limits _{1}^{4}{\sqrt {x}}dx+5\int \limits _{1}^{4}dx\\&=\left.{\frac {x^{4}}{2}}-{\frac {1}{3x}}-{\frac {8x^{\frac {3}{2}}}{3}}+5x\right|_{1}^{4}\\&=\left[{\frac {4^{4}}{2}}-{\frac {1}{3(4)}}-{\frac {8(4)^{\frac {3}{2}}}{3}}+5(4)\right]-\left[{\frac {1}{2}}-{\frac {1}{3}}-{\frac {8}{3}}+5\right]\\&=\left[128-{\frac {1}{12}}-{\frac {64}{3}}+20\right]-\left[{\frac {1}{2}}-{\frac {1}{3}}-{\frac {8}{3}}+5\right]\\&={\frac {1519}{12}}-{\frac {5}{2}}\\&={\frac {1489}{12}}\approx 124.{\overline {083}}\end{aligned}}$$

6. 已知 $f(x)={\begin{cases}x^{3},&x>1\\x+1,&x\leq 1\end{cases}}$，求 $\int \limits _{0}^{6}f(x)dx$.

根据第 4.1 节所述的端点可加性公式，我们可以根据给定的条件将积分分为两部分，对于 $f(x)=x^{3},~x>1$ 和 $f(x)=x+1,~x\leq 1$。因此，$${\begin{aligned}\int \limits _{0}^{6}f(x)dx&=\int \limits _{0}^{1}x+1dx+\int \limits _{1}^{6}x^{3}dx\\&=\left[{\frac {x^{2}}{2}}+x\right]_{0}^{1}+\left[{\frac {x^{4}}{4}}\right]_{1}^{6}\\&=\left[{\frac {1^{2}}{2}}+1\right]+\left[{\frac {6^{4}}{4}}-{\frac {1^{4}}{4}}\right]\\&={\frac {3}{2}}+{\frac {1295}{4}}\\&={\frac {1301}{4}}=325.25\end{aligned}}$$7. 令 $f(x)=\int \limits _{1}^{x}t^{2}dt$。然后求 $f'(x)$.

给定的函数可以使用微积分基本定理第一部分求导数，如第 4.2 节所述。$$f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]=x^{2}.$$这个 $x^{2}$ 从哪里来？这是从我们使用微积分基本定理第二部分找到的定积分得到的。由于我们知道导数是与反导数相反的操作，$${\begin{aligned}f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]&={\frac {d}{dx}}\left[\left.{\frac {t^{3}}{3}}\right|_{1}^{x}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}-{\frac {1^{3}}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-{\frac {d}{dx}}\left[{\frac {1}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-0\\&={\frac {1}{3}}{\frac {d}{dx}}[x^{3}]\\&={\frac {1}{3}}\cdot 3x^{3-1}\\&={\frac {1}{\cancel {3}}}\cdot {\cancel {3}}x^{2}\\&=x^{2}\end{aligned}}$$8. 给定 $A(\theta )=\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx$。求 $A'(\theta )$.

以下是定积分作为函数定义的函数的另一种形式。为了求导数，我们可以应用第 3.4 节中的微分链式法则，其中 $u=\theta ^{2}$，然后 $${\begin{aligned}A'(\theta )={\frac {d}{d\theta }}\left[\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx\right]&={\frac {d}{d\theta }}\left[\int \limits _{1}^{u}2x\cos(4x^{2})dx\right]{\frac {du}{d\theta }}\\&=\left(2u\cos(4u^{2})\right){\frac {du}{d\theta }}\\&=\left(2\theta ^{2}\cos \left(4(\theta ^{2})^{2}\right)\right){\frac {d}{d\theta }}\left[\theta ^{2}\right]\\\ &=\left(2\theta ^{2}\cos(4\theta ^{4})\right)(2\theta )\\&=4\theta ^{3}\cos(4\theta )\end{aligned}}$$9. 如果 $M(x)=\int \limits _{x^{3}}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt$。然后求 $M'(x)$.

This is quite different from the two exercise problems previously. If previously we looked at the variable as being only at one bound of the integration, then what if the variable was placed on both the upper and lower bounds? By using the additive property of a definite integral, we can break up the integral into two parts and take the derivative to generates $M'(x)$ from given function.$${\frac {d}{dx}}\left[\int \limits _{x^{3}}^{x}\cos ^{4}(t)-\sin ^{2}(\theta )dt\right]={\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt+\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]$$Therefore,$${\begin{aligned}{\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt+\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]&={\frac {d}{dx}}\left[\int \limits _{x^{3}}^{0}\cos ^{4}(t)-\sin ^{2}(t)dt\right]+{\frac {d}{dx}}\left[\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]\\&=-{\frac {d}{dx}}\left[\int \limits _{0}^{x^{3}}\cos ^{4}(t)-\sin ^{2}(t)dt\right]+{\frac {d}{dx}}\left[\int \limits _{0}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt\right]\\&=-\left({\frac {d}{du}}\left[\int \limits _{0}^{u}\cos ^{4}(t)-\sin ^{2}(t)dt\right]{\frac {du}{dx}}\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-\left(\left(\cos ^{4}(x^{3})-\sin ^{2}(x^{3})\right){\frac {d}{dx}}\left[x^{3}\right]\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-\left(\left[\cos ^{4}(x^{3})-\sin(x^{3})\right](3x^{2})\right)+\cos ^{4}(x)-\sin ^{2}(x)\\&=-3x^{2}\cos ^{4}(x^{3})+3x^{2}\sin ^{2}(x^{3})+\cos ^{4}(x)-\sin ^{2}(x)\end{aligned}}$$10. For the function${\displaystyle f(x)={\sqrt {x}}}$ over the given closed interval, ${\displaystyle [4,9]}$find the value(s) $c$ guaranteed by the mean value theorem for the definite integral.

为了找到由定积分的平均值定理保证的值 $c$，我们可以先使用下面的公式找到 $f(c)$。$$f(c)={\frac {1}{b-a}}\int \limits _{a}^{b}f(x)dx$$因此，$${\begin{aligned}f(c)&={\frac {1}{9-4}}\int \limits _{4}^{9}{\sqrt {x}}dx\\&={\frac {1}{5}}\left[{\frac {2x^{\frac {3}{2}}}{3}}\right]_{4}^{9}\\&={\frac {1}{5}}\left[{\frac {2(9)^{\frac {3}{2}}}{3}}-{\frac {2(4)^{\frac {3}{2}}}{3}}\right]\\&={\frac {1}{5}}\left[18-{\frac {16}{3}}\right]\\&={\frac {1}{5}}\left[{\frac {38}{3}}\right]\\{\sqrt {c}}&={\frac {38}{15}}\end{aligned}}$$现在，在寻找由定积分的平均值定理保证的值 c 的情况下，我们可以对等式两边平方。$${\begin{aligned}{\sqrt {c}}&={\frac {38}{15}}\\\left({\sqrt {c}}\right)^{2}&=\left({\frac {38}{15}}\right)^{2}\\c&={\frac {1444}{225}}\approx 6.{\overline {417}}\end{aligned}}$$因此，由积分平均值定理保证的值 $c$ 为 ${\frac {1444}{225}}\approx 6.{\overline {417}}$.