What is the integral from-1 to 1 of e^y-y^2-6 ?

The integral from-1 to 1 of e^y-y^2-6 is e-38/3-1/e

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integral from-1 to 1 of e^y-y^2-6

\int_{- 1}^{1} e^{y} - y^{2} - 6 d y

e - \frac{38}{3} - \frac{1}{e}

Decimal Notation

- 10.31626 \dots

Solution steps

\int_{- 1}^{1} e^{y} - y^{2} - 6 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_{- 1}^{1} e^{y} d y - \int_{- 1}^{1} y^{2} d y - \int_{- 1}^{1} 6 d y

\int_{- 1}^{1} e^{y} d y = e - \frac{1}{e}

\int_{- 1}^{1} y^{2} d y = \frac{2}{3}

\int_{- 1}^{1} 6 d y = 12

= e - \frac{1}{e} - \frac{2}{3} - 12

Combine the fractions

- 12 - \frac{2}{3} : - \frac{38}{3}

= e - \frac{38}{3} - \frac{1}{e}

\int cos^{4} (u) d u

\int sin^{5} (t) d t

\int (x^{20} - 5 x^{4})^{6} (x^{19} - x^{3}) d x

\int \frac{( x ^{2} + 1 )}{( x + 1 ) ^{2}} e^{x} d x

f (x) = x^{2} + 5 x - 3, at x = - 3

What is the integral from-1 to 1 of e^y-y^2-6 ?
The integral from-1 to 1 of e^y-y^2-6 is e-38/3-1/e