Top 50 Most Repeated APPLICATION OF INTEGRALS PYQs | JEE MAINS
A curated collection of the most important questions from APPLICATION OF INTEGRALS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from APPLICATION OF INTEGRALS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Integrate $x = y^2/4$ from $y=-4$ to $4$: $\int_{-4}^{4} \left(4 - \frac{y^2}{4}\right) dy = 32$....
Read Full Step-by-Step Solution →Area of ellipse $= \pi ab$. Here $a=4$, $b=3$, so area $= 12\pi \approx 37.70\,\text{sq units}$....
Read Full Step-by-Step Solution →Area $= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.67$. The integral gives the area under the curve above the x-axi...
Read Full Step-by-Step Solution →Using cylindrical shells: \(V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx = 2\pi\int_{0}^{4} x^{3/2}dx = \frac{128\pi}{5}\)....
Read Full Step-by-Step Solution →Split into intervals [0,2] and [2,3]; integrate piecewise linear function....
Read Full Step-by-Step Solution →Solve equations to find points of intersection; integrate difference of functions from 0 to 2....
Read Full Step-by-Step Solution →Using the disk method, $V = \pi\int_{0}^{4}(\sqrt{x})^{2}\,dx = \pi\int_{0}^{4}x\,dx = \pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=8\pi\approx25.13$....
Read Full Step-by-Step Solution →Integrate $x^2$ from 1 to 3: $\int_1^3 x^2 dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = 8.67$...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Application of Integ...
Read Full Step-by-Step Solution →Foundational check for Application of Integrals in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Integrate $x = y^2/4$ from $y=-4$ to $4$: $\int_{-4}^{4} \left(4 - \frac{y^2}{4}\right) dy = 32$....
Read Full Step-by-Step Solution →Area of ellipse $= \pi ab$. Here $a=4$, $b=3$, so area $= 12\pi \approx 37.70\,\text{sq units}$....
Read Full Step-by-Step Solution →Area $= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.67$. The integral gives the area under the curve above the x-axi...
Read Full Step-by-Step Solution →Using cylindrical shells: \(V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx = 2\pi\int_{0}^{4} x^{3/2}dx = \frac{128\pi}{5}\)....
Read Full Step-by-Step Solution →Split into intervals [0,2] and [2,3]; integrate piecewise linear function....
Read Full Step-by-Step Solution →Solve equations to find points of intersection; integrate difference of functions from 0 to 2....
Read Full Step-by-Step Solution →Using the disk method, $V = \pi\int_{0}^{4}(\sqrt{x})^{2}\,dx = \pi\int_{0}^{4}x\,dx = \pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=8\pi\approx25.13$....
Read Full Step-by-Step Solution →Integrate $x^2$ from 1 to 3: $\int_1^3 x^2 dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = 8.67$...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Application of Integ...
Read Full Step-by-Step Solution →Foundational check for Application of Integrals in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Integrate $x = y^2/4$ from $y=-4$ to $4$: $\int_{-4}^{4} \left(4 - \frac{y^2}{4}\right) dy = 32$....
Read Full Step-by-Step Solution →Area of ellipse $= \pi ab$. Here $a=4$, $b=3$, so area $= 12\pi \approx 37.70\,\text{sq units}$....
Read Full Step-by-Step Solution →Area $= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.67$. The integral gives the area under the curve above the x-axi...
Read Full Step-by-Step Solution →Using cylindrical shells: \(V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx = 2\pi\int_{0}^{4} x^{3/2}dx = \frac{128\pi}{5}\)....
Read Full Step-by-Step Solution →Split into intervals [0,2] and [2,3]; integrate piecewise linear function....
Read Full Step-by-Step Solution →Solve equations to find points of intersection; integrate difference of functions from 0 to 2....
Read Full Step-by-Step Solution →Using the disk method, $V = \pi\int_{0}^{4}(\sqrt{x})^{2}\,dx = \pi\int_{0}^{4}x\,dx = \pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=8\pi\approx25.13$....
Read Full Step-by-Step Solution →Integrate $x^2$ from 1 to 3: $\int_1^3 x^2 dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = 8.67$...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Application of Integ...
Read Full Step-by-Step Solution →Foundational check for Application of Integrals in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Integrate $x = y^2/4$ from $y=-4$ to $4$: $\int_{-4}^{4} \left(4 - \frac{y^2}{4}\right) dy = 32$....
Read Full Step-by-Step Solution →Area of ellipse $= \pi ab$. Here $a=4$, $b=3$, so area $= 12\pi \approx 37.70\,\text{sq units}$....
Read Full Step-by-Step Solution →Area $= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.67$. The integral gives the area under the curve above the x-axi...
Read Full Step-by-Step Solution →Using cylindrical shells: \(V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx = 2\pi\int_{0}^{4} x^{3/2}dx = \frac{128\pi}{5}\)....
Read Full Step-by-Step Solution →Split into intervals [0,2] and [2,3]; integrate piecewise linear function....
Read Full Step-by-Step Solution →Solve equations to find points of intersection; integrate difference of functions from 0 to 2....
Read Full Step-by-Step Solution →Using the disk method, $V = \pi\int_{0}^{4}(\sqrt{x})^{2}\,dx = \pi\int_{0}^{4}x\,dx = \pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=8\pi\approx25.13$....
Read Full Step-by-Step Solution →Integrate $x^2$ from 1 to 3: $\int_1^3 x^2 dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = 8.67$...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Application of Integ...
Read Full Step-by-Step Solution →Foundational check for Application of Integrals in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Integrate $x = y^2/4$ from $y=-4$ to $4$: $\int_{-4}^{4} \left(4 - \frac{y^2}{4}\right) dy = 32$....
Read Full Step-by-Step Solution →Area of ellipse $= \pi ab$. Here $a=4$, $b=3$, so area $= 12\pi \approx 37.70\,\text{sq units}$....
Read Full Step-by-Step Solution →Area $= \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.67$. The integral gives the area under the curve above the x-axi...
Read Full Step-by-Step Solution →Using cylindrical shells: \(V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx = 2\pi\int_{0}^{4} x^{3/2}dx = \frac{128\pi}{5}\)....
Read Full Step-by-Step Solution →Split into intervals [0,2] and [2,3]; integrate piecewise linear function....
Read Full Step-by-Step Solution →Solve equations to find points of intersection; integrate difference of functions from 0 to 2....
Read Full Step-by-Step Solution →