QUESTION IMAGE
Question
use these values to evaluate the given definite integrals.
\\( \int _ { 0 } ^ { 2 } f ( x ) d x = - 10, \quad \int _ { 0 } ^ { 2 } g ( x ) d x = 8, \quad \int _ { 0 } ^ { 3 } f ( x ) d x = 6, \quad \int _ { 2 } ^ { 3 } g ( x ) d x = 2 \\)
\\( \
\\)
Step1: Split the integral
$\int_{0}^{2} [f(x) + g(x)]dx = \int_{0}^{2}f(x)dx + \int_{0}^{2}g(x)dx$
Step2: Substitute known values
$= -10 + 8 = -2$
Step3: Split the integral
$\int_{0}^{3} [f(x) - g(x)]dx = \int_{0}^{3}f(x)dx - \int_{0}^{3}g(x)dx$
Step4: Compute $\int_{0}^{3}g(x)dx$
$\int_{0}^{3}g(x)dx = \int_{0}^{2}g(x)dx + \int_{2}^{3}g(x)dx = 8 + 2 = 10$
Step5: Substitute values
$= 6 - 10 = -4$
Step6: Split the integral
$\int_{2}^{3} [3f(x) + 2g(x)]dx = 3\int_{2}^{3}f(x)dx + 2\int_{2}^{3}g(x)dx$
Step7: Compute $\int_{2}^{3}f(x)dx$
$\int_{2}^{3}f(x)dx = \int_{0}^{3}f(x)dx - \int_{0}^{2}f(x)dx = 6 - (-10) = 16$
Step8: Substitute values
$= 3(16) + 2(2) = 48 + 4 = 52$
Step9: Split the integral
$\int_{0}^{3} [af(x) + g(x)]dx = a\int_{0}^{3}f(x)dx + \int_{0}^{3}g(x)dx = 0$
Step10: Solve for $a$
$6a + 10 = 0 \implies 6a = -10 \implies a = -\frac{10}{6} = -\frac{5}{3}$
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-2
-4
52
$-\frac{5}{3}$