What is the integral from-3 to 4 of x^2+sin(x)+e^x ?

The integral from-3 to 4 of x^2+sin(x)+e^x is 91/3-cos(4)+cos(3)+e^4-1/(e^3)

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integral from-3 to 4 of x^2+sin(x)+e^x

\int_{- 3}^{4} x^{2} + sin (x) + e^{x} d x

\frac{91}{3} - cos (4) + cos (3) + e^{4} - \frac{1}{e ^{3}}

Decimal

84.54534 \dots

Solution steps

\int_{- 3}^{4} x^{2} + sin (x) + e^{x} d x

Apply the Sum Rule:

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

= \int_{- 3}^{4} x^{2} d x + \int_{- 3}^{4} sin (x) d x + \int_{- 3}^{4} e^{x} d x

\int_{- 3}^{4} x^{2} d x = \frac{91}{3}

\int_{- 3}^{4} sin (x) d x = - cos (4) + cos (3)

\int_{- 3}^{4} e^{x} d x = e^{4} - \frac{1}{e ^{3}}

= \frac{91}{3} - cos (4) + cos (3) + e^{4} - \frac{1}{e ^{3}}

What is the integral from-3 to 4 of x^2+sin(x)+e^x ?
The integral from-3 to 4 of x^2+sin(x)+e^x is 91/3-cos(4)+cos(3)+e^4-1/(e^3)