Solution : On solving the equations x – 7y + 5 = 0 and 3x + y = 0 by using point of intersection formula, we get x = \(-5\over 22\) and y = \(15\over 22\) So, given lines intersect at \(({-5\over 22}., {15\over 22})\) Let the equation of the required line be x = […]
Solution : Inner Radius (r) = \(24\over 2\) = 12 cm Outer Radius (R) = \(28\over 2\) = 14 cm Height of Pipe = 35 cm Volume = \(\pi (R^2 – r^2) h\) = \(\pi \times 52 \times 35\) = 5720 \(cm^3\) Mass of 1 \(cm^3\) wood = 0.6 kg Mass of 5720 \(cm^3\) wood
Solution : Lateral or Curved Surface Area of Cylinder = \(2 \pi rh\) \(\implies\) \(2 \pi rh\) = 94.2 \(\implies\) \(2 \pi r \times 5\) = 94.2 \(\implies\) r = 3 cm Given, height = 5 cm Volume of Cylinder = \(\pi r^2 h\) = \(\pi \times 9 \times 5\) = \(3.14 \times 9 \times
Solution : Circumference of the base cylindrical vessel = \(2\pi r\) \(\implies\) \(2\pi r\) = 132 \(\implies\) r = 21 cm Given, height = 25 cm Volume of cylinder = \(\pi r^2 h\) = \(\pi \times {21}^2 \times 25\) = 34650 \(cm^3\) Since 1litre = 1000 \(cm^3\) Therefore, It can hold 34.65 litres of
Here you will learn what is the formula for volume of cylinder with examples. Let’s begin – What is Cylinder ? Cylinder is a solid which has both of its ends in the form of circle. Its dimensions or sides are defined in the form of the radius of the base (r) and the height
Formula for Volume of Cylinder – Definition and Examples Read More »
Solution : We have, I = \(sec^{-1}\sqrt{x}\) dx Let x = \(sec^2t\) dx = \(2sec^2 t tan t\) dt I = t.\(2sec^2 t tan t\) dt u = t and v = \(tan^2 t\) I = \(\int\) u.dv = u.v – \(\int\) v.du = \(t.tan^2 t\) – \(\int\) \(tan^2 t\) dt I = \(t.tan^2 t\)
What is the integration of sec inverse root x ? Read More »
Solution : We have, I = \(cos^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(cos^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(cos^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(cos^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(cos^{-1}\sqrt{x}\) – \(\int\) \(-1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(cos^{-1}\sqrt{x}\) –
What is the integration of cos inverse root x ? Read More »
Solution : We have, I = \(\int\) \(x cos^{-1} x\) dx By using integration by parts formula, I = \(cos^{-1} x\) \(x^2\over 2\) – \(\int\) \(-1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx I = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over
What is the integration of x cos inverse x ? Read More »
Solution : We have, I = \(\int\) \(sin^{-1}(cos x)\) dx By using integration formula, cos x = \(sin({\pi\over 2} – x)\) I = \(\int\) \(sin^{-1}[sin({\pi\over 2} – x)]\) dx I = \(\int\) \(({\pi\over 2} – x)\) dx I = \({\pi\over 2}x\) – \(x^2\over 2\) + C Similar Questions What is the integration of sin inverse
What is integration of sin inverse cos x ? Read More »
Solution : We have, I = \(sin^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(sin^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(sin^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(sin^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(sin^{-1}\sqrt{x}\) – \(\int\) \(1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(sin^{-1}\sqrt{x}\) –
What is the integration of sin inverse root x ? Read More »
