5.7-bo'lim : Aniq integrallarni hisoblash
18. Iloji bo'lsa, quyidagi integralni baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
\[\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}}\]Yechim
Ushbu integralni bajarish uchun mutlaq qiymat satrlarini "olib tashlashimiz" kerak. Biroq, buni amalga oshirish uchun biz \(4w + 3\) qayerda ijobiy va salbiy ekanligini bilishimiz kerak.
\(4w + 3\) chiziq tenglama bo'lgani uchun biz funktsiyaning quyidagi ijobiy/salbiy tabiatiga ega ekanligimiz aniq bo'lishi kerak.
\[\begin{align*}w & < - \frac{3}{4}\hspace{0.5in} \Rightarrow \hspace{0.5in} 4w + 3 < 0\\ w & > - \frac{3}{4}\hspace{0.5in} \Rightarrow \hspace{0.5in} 4w + 3 > 0\end{align*}\]2-qadam
Shunday qilib, mutlaq qiymat satrlarini olib tashlash uchun biz faqat \(w = - \frac{3}{4}\) da integralni buzishimiz kerak.
\[\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{\left| {4w + 3} \right|\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{\left| {4w + 3} \right|\,dw}}\]Shunday qilib, birinchi integralda bizda \( - 1 \le w \le - \frac{3}{4}\) bor va shuning uchun birinchi integralda \(\left| {4w + 3} \right| = - \left( {4w + 3} \right)\) mavjud. Xuddi shunday, ikkinchi integralda bizda \( - \frac{3}{4} \le w \le 0\) bor va shuning uchun ikkinchi integralda \(\left| {4w + 3} \right| = 4w + 3\) mavjud. Yoki,
\[\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{ - \left( {4w + 3} \right)\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{4w + 3\,dw}}\]3-qadam
Bu nuqtada qilishimiz kerak bo'lgan narsa - har bir integralni baholash. Mana o'sha ish.
\[\begin{align*}\int_{{ - 1}}^{0}{{\left| {4w + 3} \right|\,dw}} & = \int_{{ - 1}}^{{ - \frac{3}{4}}}{{ - 4w - 3\,dw}} + \int_{{ - \frac{3}{4}}}^{0}{{4w + 3\,dw}} = \left. {\left( { - 2{w^2} - 3w} \right)} \right|_{ - 1}^{ - \frac{3}{4}} + \left. {\left( {2{w^2} + 3w} \right)} \right|_{ - \frac{3}{4}}^0\\ & = \left[ {\frac{9}{8} - 1} \right] + \left[ {0 - \left( { - \frac{9}{8}} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}}}\end{align*}\]