If $\int_{1}^{5} f(x) , dx = 3$ and $\int_{5}^{10} f(x) , dx = 5$ , find $\int_{1}^{10} f(x) , dx$ .

Use the property $\int_{a}^{b} f(x) , dx + \int_{b}^{c} f(x) , dx = \int_{a}^{c} f(x) , dx$ . Thus, $\int_{1}^{10} f(x) , dx = 3 + 5 = 8$ .

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If $\int_{1}^{5} f(x) , dx = 3$ and $\int_{5}^{10} f(x) , dx = 5$ , find $\int_{1}^{10} f(x) , dx$ .

Use the property $\int_{a}^{b} f(x) , dx + \int_{b}^{c} f(x) , dx = \int_{a}^{c} f(x) , dx$ . Thus, $\int_{1}^{10} f(x) , dx = 3 + 5 = 8$ .

If $\int_{1}^{10} g(x) , dx = 12$ and $\int_{6}^{10} g(x) , dx = -7$ , find $\int_{1}^{6} g(x) , dx$ .

Use the property $\int_{a}^{c} f(x) , dx - \int_{b}^{c} f(x) , dx = \int_{a}^{b} f(x) , dx$ . Thus, $\int_{1}^{6} g(x) , dx = 12 - (-7) = 19$ .

If $\int_{1}^{10} f(x) , dx = 15$ and $\int_{10}^{6} f(x) , dx = 12$ , find $\int_{1}^{6} f(x) , dx$ .

First, reverse the limits: $\int_{6}^{10} f(x) , dx = -12$ . Then, $\int_{1}^{6} f(x) , dx = \int_{1}^{10} f(x) , dx - \int_{6}^{10} f(x) , dx = 15 - (-12) = 27$ .

If $\int_{1}^{19} h(x) , dx = 17$ , $\int_{6}^{19} h(x) , dx = 2$ , and $\int_{4}^{6} h(x) , dx = -3$ , find $\int_{1}^{4} h(x) , dx$ .

$\int_{1}^{4} h(x) , dx = \int_{1}^{19} h(x) , dx - \int_{6}^{19} h(x) , dx - \int_{4}^{6} h(x) , dx = 17 - 2 - (-3) = 18$ .

If $\int_{1}^{8} f(x) , dx = -8$ and $\int_{8}^{30} f(x) , dx = 200$ , find $\int_{1}^{30} f(x) , dx$ .

$\int_{1}^{30} f(x) , dx = \int_{1}^{8} f(x) , dx + \int_{8}^{30} f(x) , dx = -8 + 200 = 192$ .

If $\int_{1}^{4} g(x) , dx = -8$ and $\int_{4}^{2} g(x) , dx = 3$ , find $\int_{1}^{2} g(x) , dx$ .

First, reverse the limits: $\int_{2}^{4} g(x) , dx = -3$ . Then, $\int_{1}^{2} g(x) , dx = \int_{1}^{4} g(x) , dx - \int_{2}^{4} g(x) , dx = -8 - (-3) = -5$ .

What is the Zero Rule for definite integrals?

$\int_{a}^{a} f(x) , dx = 0$

What is the formula for reversing limits of integration?

$\int_{b}^{a} f(x) , dx = -\int_{a}^{b} f(x) , dx$

How do you handle a constant multiple inside a definite integral?

$\int_{a}^{b} k cdot f(x) , dx = k \int_{a}^{b} f(x) , dx$

How do you integrate a sum or difference of functions?

$\int_{a}^{b} [f(x) pm g(x)] , dx = \int_{a}^{b} f(x) , dx pm \int_{a}^{b} g(x) , dx$

State the formula for splitting an interval of integration.

$\int_{a}^{b} f(x) , dx + \int_{b}^{c} f(x) , dx = \int_{a}^{c} f(x) , dx$

Explain why $\int_{a}^{a} f(x) , dx = 0$ .

The integral represents the area under the curve. If the upper and lower limits are the same, the 'width' of the area is zero, resulting in zero area.

Explain the concept of reversing the limits of integration.

Reversing the limits changes the direction of integration, thus negating the value of the integral because the area is now considered 'negative'.

Why can a constant be moved outside the integral?

Multiplying a function by a constant scales its area by the same factor. Thus, the constant can be factored out.

Explain the concept of splitting the interval of integration.

The total area under a curve from a to c can be found by summing the area from a to b and then from b to c.