1. 利用定积分定义计算积分$\int_{a}^{b} x dx $
A: $\frac{1}{2}(b^2 -a^2)$
B: $\frac{1}{2}$
C: $\frac{1}{2}b^2 $
D: $\frac{1}{2}(a^2 - b^2)$
A: $\frac{1}{2}(b^2 -a^2)$
B: $\frac{1}{2}$
C: $\frac{1}{2}b^2 $
D: $\frac{1}{2}(a^2 - b^2)$
举一反三
- 利用定积分定义计算积分$\int_{a}^{b} x dx $ A: $\frac{1}{2}(b^2 -a^2)$ B: $\frac{1}{2}$ C: $\frac{1}{2}b^2 $ D: $\frac{1}{2}(a^2 - b^2)$
- 积分\(\int_0^1 (x\sin\frac{1}{x^2} - \frac{1}{x}\cos\frac{1}{x^2})dx\) (不计算积分, 由判别法直接判断)
- 积分$\int_0^1 x \arctan xdx=$()。 A: $\frac{\pi}{4}+\frac{1}{2}$ B: $\frac{\pi}{4}$ C: $\frac{\pi}{4}-\frac{1}{2}$ D: $\frac{1}{2}$
- For the integral $\int_0^{+\infty}\frac{dx}{(x^2+p^2)(x^2+q^2)}$, which of the following statements are CORRECT? A: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2},p>0 \ q>0;$ B: $\frac{1}{q^2-p^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ -q>0;$ C: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2}, p>0 \ -q>0;$ D: $\frac{1}{p^2-q^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ q>0.$
- 设`\n`阶方阵`\A`满足`\|A| = 2`,则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于( ) A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]