What is the integral from x to y of (1+y^2) ?

The integral from x to y of (1+y^2) is y-x+(y^3-x^3)/3

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integral from x to y of (1+y^2)

\int_{x}^{y} (1 + y^{2}) d y

y - x + \frac{y ^{3} - x ^{3}}{3}

Solution steps

\int_{x}^{y} 1 + y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x}^{y} 1 d y + \int_{x}^{y} y^{2} d y

\int_{x}^{y} 1 d y = y - x

\int_{x}^{y} y^{2} d y = \frac{y ^{3} - x ^{3}}{3}

= y - x + \frac{y ^{3} - x ^{3}}{3}

\int_{0}^{4} \frac{x ^{2}}{x + 4} d x

\int_{1}^{2} 4 x^{3} e^{x^{4}} d x

\int_{0}^{r} π (r^{2} - x^{2}) d x

\int_{0}^{2 π} 1 cos^{2} (x) d x

\int_{- a}^{a} 4 d x

What is the integral from x to y of (1+y^2) ?
The integral from x to y of (1+y^2) is y-x+(y^3-x^3)/3