What is the integral from 1 to 3 of h(x) ?

The integral from 1 to 3 of h(x) is h*4

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integral from 1 to 3 of h(x)

\int_{1}^{3} h (x) d x

h \cdot 4

Solution steps

\int_{1}^{3} h x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= h \cdot \int_{1}^{3} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{3}

= h [\frac{x ^{2}}{2}]_{1}^{3}

Compute the boundaries:

4

= h \cdot 4

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

\int_{4}^{x} \frac{1}{1 + t + t ^{2}} d t

\int_{0}^{1} e^{6 y} y d y

\int_{- π}^{π} cos^{2} (\frac{x}{2}) d x

\int_{0}^{\frac{4 π}{3}} tan (\frac{x}{4}) d x

What is the integral from 1 to 3 of h(x) ?
The integral from 1 to 3 of h(x) is h*4