微積分/微積分基本定理/解答

1. 計算 $\int _{0}^{1}x^{6}dx$。將你的答案與你在第 4.1 節練習 1 中得到的答案進行比較。

$\int _{0}^{1}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{0}^{1}={\frac {1^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {1}{7}}=0.{\overline {142857}}}$
這與我們在第 4.1 節練習 1 中計算出的邊界一致。

2. 計算 $\int _{1}^{2}x^{6}dx$。將你的答案與你在第 4.1 節練習 2 中得到的答案進行比較。

$\int _{1}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{1}^{2}={\frac {2^{7}}{7}}-{\frac {1^{7}}{7}}={\frac {128}{7}}-{\frac {1}{7}}=\mathbf {{\frac {127}{7}}=18.{\overline {142857}}}$
這與我們在第 4.1 節練習 2 中計算出的邊界一致。

3. 計算 $\int _{0}^{2}x^{6}dx$。將你的答案與你在第 4.1 節練習 4 中得到的答案進行比較。

$\int _{0}^{2}x^{6}dx={\frac {x^{7}}{7}}{\biggr |}_{0}^{2}={\frac {2^{7}}{7}}-{\frac {0^{7}}{7}}=\mathbf {{\frac {128}{7}}=18.{\overline {285714}}}$
這與我們在第 4.1 節練習 4 中計算出的邊界一致。

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

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

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

根據第 4.1 節所述的端點可加性公式，我們可以根據給定的條件將積分分為兩部分，對於 $f(x)=x^{3},~x>1$ 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 $f(x)=\int \limits _{1}^{x}t^{2}dt$。然後求 $f'(x)$.

給定的函式可以使用微積分基本定理第一部分求導數，如第 4.2 節所述。$$f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]=x^{2}.$$這個 $x^{2}$ 從哪裡來？這是從我們使用微積分基本定理第二部分找到的定積分得到的。由於我們知道導數是與反導數相反的操作，$${\begin{aligned}f'(x)={\frac {d}{dx}}\left[\int \limits _{1}^{x}t^{2}dt\right]&={\frac {d}{dx}}\left[\left.{\frac {t^{3}}{3}}\right|_{1}^{x}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}-{\frac {1^{3}}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-{\frac {d}{dx}}\left[{\frac {1}{3}}\right]\\&={\frac {d}{dx}}\left[{\frac {x^{3}}{3}}\right]-0\\&={\frac {1}{3}}{\frac {d}{dx}}[x^{3}]\\&={\frac {1}{3}}\cdot 3x^{3-1}\\&={\frac {1}{\cancel {3}}}\cdot {\cancel {3}}x^{2}\\&=x^{2}\end{aligned}}$$8. 給定 $A(\theta )=\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx$。求 $A'(\theta )$.

以下是定積分作為函式定義的函式的另一種形式。為了求導數，我們可以應用第 3.4 節中的微分鏈式法則，其中 $u=\theta ^{2}$，然後 $${\begin{aligned}A'(\theta )={\frac {d}{d\theta }}\left[\int \limits _{1}^{\theta ^{2}}2x\cos(4x^{2})dx\right]&={\frac {d}{d\theta }}\left[\int \limits _{1}^{u}2x\cos(4x^{2})dx\right]{\frac {du}{d\theta }}\\&=\left(2u\cos(4u^{2})\right){\frac {du}{d\theta }}\\&=\left(2\theta ^{2}\cos \left(4(\theta ^{2})^{2}\right)\right){\frac {d}{d\theta }}\left[\theta ^{2}\right]\\\ &=\left(2\theta ^{2}\cos(4\theta ^{4})\right)(2\theta )\\&=4\theta ^{3}\cos(4\theta )\end{aligned}}$$9. 如果 $M(x)=\int \limits _{x^{3}}^{x}\cos ^{4}(t)-\sin ^{2}(t)dt$。然後求 $M'(x)$.

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.

為了找到由定積分的平均值定理保證的值 $c$，我們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，$${\begin{aligned}f(c)&={\frac {1}{9-4}}\int \limits _{4}^{9}{\sqrt {x}}dx\\&={\frac {1}{5}}\left[{\frac {2x^{\frac {3}{2}}}{3}}\right]_{4}^{9}\\&={\frac {1}{5}}\left[{\frac {2(9)^{\frac {3}{2}}}{3}}-{\frac {2(4)^{\frac {3}{2}}}{3}}\right]\\&={\frac {1}{5}}\left[18-{\frac {16}{3}}\right]\\&={\frac {1}{5}}\left[{\frac {38}{3}}\right]\\{\sqrt {c}}&={\frac {38}{15}}\end{aligned}}$$現在，在尋找由定積分的平均值定理保證的值 c 的情況下，我們可以對等式兩邊平方。$${\begin{aligned}{\sqrt {c}}&={\frac {38}{15}}\\\left({\sqrt {c}}\right)^{2}&=\left({\frac {38}{15}}\right)^{2}\\c&={\frac {1444}{225}}\approx 6.{\overline {417}}\end{aligned}}$$因此，由積分平均值定理保證的值 $c$ 為 ${\frac {1444}{225}}\approx 6.{\overline {417}}$.