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Questions 1 - 10

$\int_0^4-8x^2+15=?$

$-110\tfrac{2}{3}$

$230\tfrac{2}{3}$

$-230\tfrac{2}{3}$

Explanation

Use the Fundamental Theorem of Calculus. If , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore we need to find the indefinite integral.

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

Plug that back into FTOC:

$\int_a^b -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}b^3+15b+c)-(-\frac{8}{3}a^3+15a+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(4)^3+15(4))-(-\frac{8}{3}(0)^3+15(0))$ $\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(64)+(60))-(0+0)$

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-170\tfrac{2}{3}+(60))$

$\int_0^4 -8x^2+15{\mathrm{d} x}=-110\tfrac{2}{3}$

$\int_0^4-8x^2+15=?$

$-110\tfrac{2}{3}$

$230\tfrac{2}{3}$

$-230\tfrac{2}{3}$

Explanation

Use the Fundamental Theorem of Calculus. If , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore we need to find the indefinite integral.

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

Plug that back into FTOC:

$\int_a^b -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}b^3+15b+c)-(-\frac{8}{3}a^3+15a+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(4)^3+15(4))-(-\frac{8}{3}(0)^3+15(0))$ $\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(64)+(60))-(0+0)$

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-170\tfrac{2}{3}+(60))$

$\int_0^4 -8x^2+15{\mathrm{d} x}=-110\tfrac{2}{3}$

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

$2.35\cdot 10^{9}$

$2.35\cdot 10^{17}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{15}$

$2.35\cdot 10^{19}$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our , the power rule really doesn't help us. is the only function that is it's OWN anti-derivative. That means we're still going to be working with .

$\int e^x{\mathrm{d} x}=e^x+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(e^b+c)-(e^a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

$\int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sqrt{x}$ , we can use the power rule, if we turn it into an exponent:

$f(x)=\sqrt{x}=x^{\frac{1}{2}}$

This means that:

$\int f(x){\mathrm{d} x}=\int x^\frac{1}{2}{\mathrm{d} x}=\frac{2}{3}x^\frac{3}{2}+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{2}{3}b^\frac{3}{2}+c)-(\frac{2}{3}a^\frac{3}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{5}f(x){\mathrm{d} x}=(\frac{2}{3}(5)^\frac{3}{2})-(\frac{2}{3}(2)^\frac{3}{2})$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx5.56$

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

$2.35\cdot 10^{9}$

$2.35\cdot 10^{17}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{15}$

$2.35\cdot 10^{19}$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our , the power rule really doesn't help us. is the only function that is it's OWN anti-derivative. That means we're still going to be working with .

$\int e^x{\mathrm{d} x}=e^x+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(e^b+c)-(e^a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

$\int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sqrt{x}$ , we can use the power rule, if we turn it into an exponent:

$f(x)=\sqrt{x}=x^{\frac{1}{2}}$

This means that:

$\int f(x){\mathrm{d} x}=\int x^\frac{1}{2}{\mathrm{d} x}=\frac{2}{3}x^\frac{3}{2}+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{2}{3}b^\frac{3}{2}+c)-(\frac{2}{3}a^\frac{3}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{5}f(x){\mathrm{d} x}=(\frac{2}{3}(5)^\frac{3}{2})-(\frac{2}{3}(2)^\frac{3}{2})$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx5.56$

$\int_0^4-8x^2+15=?$

$-110\tfrac{2}{3}$

$230\tfrac{2}{3}$

$-230\tfrac{2}{3}$

Explanation

Use the Fundamental Theorem of Calculus. If , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore we need to find the indefinite integral.

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

Plug that back into FTOC:

$\int_a^b -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}b^3+15b+c)-(-\frac{8}{3}a^3+15a+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(4)^3+15(4))-(-\frac{8}{3}(0)^3+15(0))$ $\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(64)+(60))-(0+0)$

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-170\tfrac{2}{3}+(60))$

$\int_0^4 -8x^2+15{\mathrm{d} x}=-110\tfrac{2}{3}$

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

$2.35\cdot 10^{9}$

$2.35\cdot 10^{17}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{15}$

$2.35\cdot 10^{19}$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our , the power rule really doesn't help us. is the only function that is it's OWN anti-derivative. That means we're still going to be working with .

$\int e^x{\mathrm{d} x}=e^x+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(e^b+c)-(e^a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

$\int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?$

Explanation

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sqrt{x}$ , we can use the power rule, if we turn it into an exponent:

$f(x)=\sqrt{x}=x^{\frac{1}{2}}$

This means that:

$\int f(x){\mathrm{d} x}=\int x^\frac{1}{2}{\mathrm{d} x}=\frac{2}{3}x^\frac{3}{2}+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{2}{3}b^\frac{3}{2}+c)-(\frac{2}{3}a^\frac{3}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{5}f(x){\mathrm{d} x}=(\frac{2}{3}(5)^\frac{3}{2})-(\frac{2}{3}(2)^\frac{3}{2})$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx5.56$

What is the indefinite integral of ?

$-\frac{8}{3}x^3+15x+c$

$-\frac{8}{3}x^3+15x$

$\frac{8}{3}x^3-15x+c$

Explanation

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$