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  2. AP Calculus
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What is a definite integral?

The value obtained by integrating a function over a specific interval [a, b], representing the area under the curve.

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What is a definite integral?

The value obtained by integrating a function over a specific interval [a, b], representing the area under the curve.

Define the upper limit of integration.

The value 'b' in the definite integral ∫abf(x),dx\int_{a}^{b} f(x) , dx∫ab​f(x),dx, representing the upper bound of the interval.

Define the lower limit of integration.

The value 'a' in the definite integral ∫abf(x),dx\int_{a}^{b} f(x) , dx∫ab​f(x),dx, representing the lower bound of the interval.

What is the integrand?

The function f(x)f(x)f(x) being integrated in a definite integral.

If ∫15f(x),dx=3\int_{1}^{5} f(x) , dx = 3∫15​f(x),dx=3 and ∫510f(x),dx=5\int_{5}^{10} f(x) , dx = 5∫510​f(x),dx=5, find ∫110f(x),dx\int_{1}^{10} f(x) , dx∫110​f(x),dx.

Use the property ∫abf(x),dx+∫bcf(x),dx=∫acf(x),dx\int_{a}^{b} f(x) , dx + \int_{b}^{c} f(x) , dx = \int_{a}^{c} f(x) , dx∫ab​f(x),dx+∫bc​f(x),dx=∫ac​f(x),dx. Thus, ∫110f(x),dx=3+5=8\int_{1}^{10} f(x) , dx = 3 + 5 = 8∫110​f(x),dx=3+5=8.

If ∫110g(x),dx=12\int_{1}^{10} g(x) , dx = 12∫110​g(x),dx=12 and ∫610g(x),dx=−7\int_{6}^{10} g(x) , dx = -7∫610​g(x),dx=−7, find ∫16g(x),dx\int_{1}^{6} g(x) , dx∫16​g(x),dx.

Use the property ∫acf(x),dx−∫bcf(x),dx=∫abf(x),dx\int_{a}^{c} f(x) , dx - \int_{b}^{c} f(x) , dx = \int_{a}^{b} f(x) , dx∫ac​f(x),dx−∫bc​f(x),dx=∫ab​f(x),dx. Thus, ∫16g(x),dx=12−(−7)=19\int_{1}^{6} g(x) , dx = 12 - (-7) = 19∫16​g(x),dx=12−(−7)=19.

If ∫110f(x),dx=15\int_{1}^{10} f(x) , dx = 15∫110​f(x),dx=15 and ∫106f(x),dx=12\int_{10}^{6} f(x) , dx = 12∫106​f(x),dx=12, find ∫16f(x),dx\int_{1}^{6} f(x) , dx∫16​f(x),dx.

First, reverse the limits: ∫610f(x),dx=−12\int_{6}^{10} f(x) , dx = -12∫610​f(x),dx=−12. Then, ∫16f(x),dx=∫110f(x),dx−∫610f(x),dx=15−(−12)=27\int_{1}^{6} f(x) , dx = \int_{1}^{10} f(x) , dx - \int_{6}^{10} f(x) , dx = 15 - (-12) = 27∫16​f(x),dx=∫110​f(x),dx−∫610​f(x),dx=15−(−12)=27.

If ∫119h(x),dx=17\int_{1}^{19} h(x) , dx = 17∫119​h(x),dx=17, ∫619h(x),dx=2\int_{6}^{19} h(x) , dx = 2∫619​h(x),dx=2, and ∫46h(x),dx=−3\int_{4}^{6} h(x) , dx = -3∫46​h(x),dx=−3, find ∫14h(x),dx\int_{1}^{4} h(x) , dx∫14​h(x),dx.

∫14h(x),dx=∫119h(x),dx−∫619h(x),dx−∫46h(x),dx=17−2−(−3)=18\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∫14​h(x),dx=∫119​h(x),dx−∫619​h(x),dx−∫46​h(x),dx=17−2−(−3)=18.

If ∫18f(x),dx=−8\int_{1}^{8} f(x) , dx = -8∫18​f(x),dx=−8 and ∫830f(x),dx=200\int_{8}^{30} f(x) , dx = 200∫830​f(x),dx=200, find ∫130f(x),dx\int_{1}^{30} f(x) , dx∫130​f(x),dx.

∫130f(x),dx=∫18f(x),dx+∫830f(x),dx=−8+200=192\int_{1}^{30} f(x) , dx = \int_{1}^{8} f(x) , dx + \int_{8}^{30} f(x) , dx = -8 + 200 = 192∫130​f(x),dx=∫18​f(x),dx+∫830​f(x),dx=−8+200=192.

If ∫14g(x),dx=−8\int_{1}^{4} g(x) , dx = -8∫14​g(x),dx=−8 and ∫42g(x),dx=3\int_{4}^{2} g(x) , dx = 3∫42​g(x),dx=3, find ∫12g(x),dx\int_{1}^{2} g(x) , dx∫12​g(x),dx.

First, reverse the limits: ∫24g(x),dx=−3\int_{2}^{4} g(x) , dx = -3∫24​g(x),dx=−3. Then, ∫12g(x),dx=∫14g(x),dx−∫24g(x),dx=−8−(−3)=−5\int_{1}^{2} g(x) , dx = \int_{1}^{4} g(x) , dx - \int_{2}^{4} g(x) , dx = -8 - (-3) = -5∫12​g(x),dx=∫14​g(x),dx−∫24​g(x),dx=−8−(−3)=−5.

What is the Zero Rule for definite integrals?

∫aaf(x),dx=0\int_{a}^{a} f(x) , dx = 0∫aa​f(x),dx=0

What is the formula for reversing limits of integration?

∫baf(x),dx=−∫abf(x),dx\int_{b}^{a} f(x) , dx = -\int_{a}^{b} f(x) , dx∫ba​f(x),dx=−∫ab​f(x),dx

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

∫abkcdotf(x),dx=k∫abf(x),dx\int_{a}^{b} k cdot f(x) , dx = k \int_{a}^{b} f(x) , dx∫ab​kcdotf(x),dx=k∫ab​f(x),dx

How do you integrate a sum or difference of functions?

∫ab[f(x)pmg(x)],dx=∫abf(x),dxpm∫abg(x),dx\int_{a}^{b} [f(x) pm g(x)] , dx = \int_{a}^{b} f(x) , dx pm \int_{a}^{b} g(x) , dx∫ab​[f(x)pmg(x)],dx=∫ab​f(x),dxpm∫ab​g(x),dx

State the formula for splitting an interval of integration.

∫abf(x),dx+∫bcf(x),dx=∫acf(x),dx\int_{a}^{b} f(x) , dx + \int_{b}^{c} f(x) , dx = \int_{a}^{c} f(x) , dx∫ab​f(x),dx+∫bc​f(x),dx=∫ac​f(x),dx