87593 The solution of \(\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0\) is

87595 Cube root of 18 by using Newton- Raphson method will be

87598 A curve is drawn to pass through the points given by the following table.
| $\mathbf{x}$ | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{y}$ | 2 | 2.4 | 2.7 | 2.8 | 3 | 2.6 | 2.1 |
Using Simpson's 1/3rd rule, estimate the area bounded by the curve, the \(x\)-axis and the lines \(\mathrm{x}=\mathbf{1}, \mathrm{x}=4\)

87600 Calculate by Trapezoidal rule an approximate value of \(\int_{-3}^{3} x^{4}\) by taking seven equidistant ordinates

87593 The solution of \(\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0\) is

87595 Cube root of 18 by using Newton- Raphson method will be

87598 A curve is drawn to pass through the points given by the following table.
| $\mathbf{x}$ | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{y}$ | 2 | 2.4 | 2.7 | 2.8 | 3 | 2.6 | 2.1 |
Using Simpson's 1/3rd rule, estimate the area bounded by the curve, the \(x\)-axis and the lines \(\mathrm{x}=\mathbf{1}, \mathrm{x}=4\)

87600 Calculate by Trapezoidal rule an approximate value of \(\int_{-3}^{3} x^{4}\) by taking seven equidistant ordinates

87593 The solution of \(\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0\) is

87595 Cube root of 18 by using Newton- Raphson method will be

87598 A curve is drawn to pass through the points given by the following table.
| $\mathbf{x}$ | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{y}$ | 2 | 2.4 | 2.7 | 2.8 | 3 | 2.6 | 2.1 |
Using Simpson's 1/3rd rule, estimate the area bounded by the curve, the \(x\)-axis and the lines \(\mathrm{x}=\mathbf{1}, \mathrm{x}=4\)

87600 Calculate by Trapezoidal rule an approximate value of \(\int_{-3}^{3} x^{4}\) by taking seven equidistant ordinates

87593 The solution of \(\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0\) is

87595 Cube root of 18 by using Newton- Raphson method will be

87598 A curve is drawn to pass through the points given by the following table.
| $\mathbf{x}$ | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{y}$ | 2 | 2.4 | 2.7 | 2.8 | 3 | 2.6 | 2.1 |
Using Simpson's 1/3rd rule, estimate the area bounded by the curve, the \(x\)-axis and the lines \(\mathrm{x}=\mathbf{1}, \mathrm{x}=4\)

87600 Calculate by Trapezoidal rule an approximate value of \(\int_{-3}^{3} x^{4}\) by taking seven equidistant ordinates