Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
124 views
in Limits by (415 points)
Solve ∫y=0^{1}∫x=y^{2}^{1}∫z=0^{1-x} x dzdxdy

Please log in or register to answer this question.

1 Answer

+2 votes
by (710 points)

\(\displaystyle\int\limits^1_{y=0} \) \(\displaystyle\int\limits^1_{\mathrm x=y^2}\) \(\displaystyle\int\limits^{1-\mathrm x}_{z=0}\) x dz dx dy

\(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x \,\left[z\right]^{1-\mathrm x}_0d\mathrm x\right)dy\)

\(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x(1-\mathrm x)d\mathrm x\right)dy\)

\(=\displaystyle\int\limits^1_0\left(\frac{\mathrm x^2}{2}-\frac{\mathrm x^3}{3}\right)^1_{y^2}dy\)

\(=\displaystyle\int\limits^1_0\left(\frac{1}{2}(1-y^4)-\frac{1}{3}(1-y^6)\right)dy\)

\(=\frac{1}{2}\displaystyle\int\limits^1_0(1-y^4)dy\) \(-\frac{1}{3}\displaystyle\int\limits^1_0(1-y^6)dy\)

\(=\frac{1}{2}\left[y-\frac{y^5}{5}\right]^1_0\) \(-\frac{1}{3}\left[y-\frac{y^7}{7}\right]^1_0\)

\(=\frac{1}{2}\left(1-\frac{1}{5}\right)-\frac{1}{3}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}\times \frac{4}{5}-\frac{1}{3}\times \frac{6}{7}\)

\(=\frac{2}{5}-\frac{2}{7}\) \(=\frac{14-10}{35}=\frac{4}{35}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...