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Notes

Fundamental Theorem of Calculus

\(\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt \,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)

Questions & Solutions

\(\textbf{1)}\) \(\frac{d}{dx}\left(\displaystyle\int_3^x 3t^3+4t\,dt\right)\)

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\(\text{The answer is } 3x^3+4x\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_3^x 3t^3+4t\,dt\right)\,\,=\,\, \left[\left(3x^3+4x\right)\cdot (1)\right]-\left[\left(3(3)^3+4(3)\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } 3x^3+4x\)

\(\textbf{2)}\) \(\frac{d}{dx}\left(\displaystyle\int_x^3\sin \left(t\right)\,dt\right)\)

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\(\text{The answer is } -\sin \left(x\right)\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int _x^3\sin \left(t\right)\,dt\right)\,\,=\,\, \left[\left(\sin \left(3\right)\right)\cdot (0)\right]-\left[\left(\sin \left(x\right)\right) \cdot (1)\right] \)


\(\,\,\,\,\,\,\text{The answer is } -\sin \left(x\right)\)

\(\textbf{3)}\) \(\frac{d}{dx}\left(\displaystyle\int_3^{x^2}3t+5\,dt\right)\)

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\(\text{The answer is } 6x^3+10x\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_3^{x^2}3t+5\,dt\right)\,\,=\,\, \left[\left(3x^2+5\right)\cdot (2x)\right]-\left[\left(3(3)+5\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } 6x^3+10x\)

\(\textbf{4)}\) \(\frac{d}{dx}\left(\displaystyle\int _x^{2x}te^t\,dt\right)\)

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\(\text{The answer is } 4e^{2x}x-xe^x \)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int _x^{2x}te^t\,dt\right)\,\,=\,\, \left[\left((2x)e^{(2x)}\right)\cdot (2)\right]-\left[\left((x)e^{(x)}\right) \cdot (1)\right] \)


\(\,\,\,\,\,\,\text{The answer is } 4e^{2x}x-xe^x\)

\(\textbf{5)}\) \(\frac{d}{dx}\left(\displaystyle\int _6^x e^{3t}\sin\left(te^{4t}-\tan^2\left(t\right)\right)\right)\,dt\)

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\(\text{The answer is } e^{3x}\sin\left(xe^{4x}-\tan^2\left(x\right)\right)\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int _6^x e^{3t}\sin\left(te^{4t}-\tan^2\left(t\right)\right)\right)\,dt\,\,=\,\, \left[\left(e^{3x}\sin\left(xe^{4x}-\tan^2\left(x\right)\right)\right)\cdot (1)\right]-\left[\left(e^{3(6)}\sin\left((6)e^{4(6)}-\tan^2\left(6\right)\right)\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } e^{3x}\sin\left(xe^{4x}-\tan^2\left(x\right)\right)\)

\(\textbf{6)}\) \(\frac{d}{dx}\left(\displaystyle\int_{\sqrt{x}}^{x^2}t^2\,dt\right)\)

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\(\text{The answer is } 2x^5-\displaystyle \frac{\sqrt{x}}{2}\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_{\sqrt{x}}^{x^2}t^2\,dt\right)\,\,=\,\, \left[\left((x^2)^2\right)\cdot (2x)\right]-\left[\left((\sqrt{x})^2\right) \cdot \left(\frac{1}{2\sqrt{x}}\right)\right] \)


\(\,\,\,\,\,\,\text{The answer is } 2x^5-\displaystyle \frac{\sqrt{x}}{2}\)

\(\textbf{7)}\) \(\frac{d}{dx}\left(\displaystyle\int_9^{x} t \cos^2(t)\left(t^4-t^3\right)\,dt \right)\)

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\(\text{The answer is } x \cos^2(x)\left(x^4-x^3\right)\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_9^{x} t \cos^2(t)\left(t^4-t^3\right)\,dt\right)\,\,=\,\, \left[\left(x \cos^2(x)\right)\left(x^4-x^3\right)\cdot (1)\right]-\left[\left((9) \cos^2(9)\right)\left((9)^4-(9)^3\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } x \cos^2(x)\left(x^4-x^3\right)\)

\(\textbf{8)}\) \(\frac{d}{dx}\left(\displaystyle\int_0^{\cos \left(x\right)}t^2 \,dt \right)\)

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\(\text{The answer is } -\cos ^2\left(x\right)\sin \left(x\right)\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_0^{\cos \left(x\right)}t^2 \,dt \right)\,\,=\,\, \left[\left(\left(\cos(x)\right)^2\right)\cdot (-\sin(x))\right]-\left[\left((0)^2\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } -\cos ^2\left(x\right)\sin \left(x\right)\)

\(\textbf{9)}\) \(\frac{d}{dx}\left(\displaystyle\int _{5x^2}^7\frac{t^2}{t-5} \,dt\right)\)

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\(\text{The answer is } -\displaystyle \frac{50x^5}{x^2-1}\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int _{5x^2}^7\frac{t^2}{t-5} \,dt\right)\,\,=\,\, \left[\left(\frac{(7)^2}{(7)-5}\right)\cdot (0)\right]-\left[\left(\frac{t^2}{t-5}\right) \cdot (10x)\right] \)


\(\,\,\,\,\,\,\text{The answer is } -\displaystyle \frac{50x^5}{x^2-1}\)

\(\textbf{10)}\) \(\frac{d}{dx}\left(\displaystyle\int_1^{3}t^3 \, dt\right)\)

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\(\text{The answer is } 0\)

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\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right]\)


\(\,\,\,\,\,\,\frac{d}{dx}\left(\displaystyle\int_1^{3}t^3 \, dt\right)\,\,=\,\, \left[\left((3)^3\right)\cdot (0)\right]-\left[\left((1)^3\right) \cdot (0)\right] \)


\(\,\,\,\,\,\,\text{The answer is } 0\)

\(\textbf{11)}\) Find \( F'(x) \) if \(F(x)=\displaystyle\int^{\pi/4}_{\sqrt{x}}t \tan{(t)} \,dt\)

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\(F'(x)=\frac{d}{dx}\displaystyle\int^{\pi/4}_{\sqrt{x}}t \tan{(t)} \,dt= \frac{\left(\sqrt{x}\right)\tan{\left(\sqrt{x}\right)}}{2\sqrt{x}} \)

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\(\,\,\,\,\,\,F(x)=\displaystyle\int^{\pi/4}_{\sqrt{x}}t \tan{(t)} \,dt\)


\(\,\,\,\,\,\,\frac{d}{dx}\displaystyle\int_{f(x)}^{g(x)} h(t) \, dt\,\,=\,\, \left[h\left(g(x)\right)\cdot g'(x)\right]-\left[h\left(f(x)\right) \cdot f'(x)\right] \)


\(\,\,\,\,\,\,F'(x)=\frac{d}{dx}\displaystyle\int^{\pi/4}_{\sqrt{x}}t \tan{(t)} \,dt=\,\, \left[\left( \pi/4\right)\tan{\left(\pi/4\right)}\cdot 0\right]-\left[\left(\sqrt{x}\right)\tan{\left(\sqrt{x}\right)} \cdot \left(\frac{-1}{2\sqrt{x}} \right )\right] \)


\(\,\,\,\,\,\,F'(x)=\frac{d}{dx}\displaystyle\int^{\pi/4}_{\sqrt{x}}t \tan{(t)} \,dt= \frac{\left(\sqrt{x}\right)\tan{\left(\sqrt{x}\right)}}{2\sqrt{x}} \)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Trapezoidal Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{b-a}{2n}\left[f(a)+2f(x_1)+2f(x_2)+…+2fx_{n-1}+f(b)\right]…\)

\(\bullet\text{ Properties of Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}cf(x) \, dx=c\displaystyle \int_{a}^{b}f(x) \,dx…\)

\(\bullet\text{ Indefinite Integrals- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int x^n \, dx = \displaystyle \frac{x^{n+1}}{n+1}+C…\)

\(\bullet\text{ Indefinite Integrals- Trig Functions}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int \cos{x} \, dx=\sin{x}+C…\)

\(\bullet\text{ Definite Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{5}^{7} x^3 \, dx…\)

\(\bullet\text{ Integration by Substitution}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int (x^2+3)^3(2x) \,dx…\)

\(\bullet\text{ Area of Region Between Two Curves}\)
\(\,\,\,\,\,\,\,\,A=\displaystyle \int_{a}^{b}\left[f(x)-g(x)\right]\,dx…\)

\(\bullet\text{ Arc Length}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}\sqrt{1+\left[f'(x)\right]^2} \,dx…\)

\(\bullet\text{ Average Function Value}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{1}{b-a} \int_{a}^{b}f(x) \,dx\)

\(\bullet\text{ Volume by Cross Sections}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Disk Method}\)
\(\,\,\,\,\,\,\,\,V=\displaystyle \int_{a}^{b}\left[f(x)\right]^2\,dx…\)

\(\bullet\text{ Cylindrical Shells}\)
\(\,\,\,\,\,\,\,\,V=2 \pi \displaystyle \int_{a}^{b} y f(y) \, dy…\)

In Summary

The Fundamental Theorem of Calculus establishes a link between the concept of a derivative and the concept of an integral. Essentially, it allows us to use the tools of calculus to solve problems involving the accumulation of quantities.

More specifically, the Fundamental Theorem of Calculus states that if f is a continuous function defined on a closed interval [a,b], and if F is a function defined on that interval such that F'(x) = f(x) for all x in [a,b], then the definite integral of f from a to b is equal to F(b) – F(a).

We learn about the Fundamental Theorem of Calculus because it is a crucial result that allows us to solve a wide range of problems in mathematics and science. It is used in many different fields, including physics, engineering, economics, and more.

The Fundamental Theorem of Calculus is typically covered in a Calculus I or Calculus II course, depending on the curriculum of the school. It is an important topic in both of these classes and is often used as a basis for more advanced concepts in calculus.

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